AP Chemistry Unit 4: Chemical Reactions — Free Review Games.

The coefficients show 2 moles of \(H_2\) produce 2 moles of \(H_2O\), giving a \(2:2\) or simplified \(1:1\) mole ratio.

Two reactants (\(Mg\) and \(O_2\)) combine to form a single product ($MgO$), which defines a synthesis or combination reaction.

The law of conservation of mass requires that atoms are neither created nor destroyed. A balanced equation has equal numbers of each type of atom on both sides.

NaCl is an ionic compound that completely dissociates into Na+ and Cl- ions in water, making it a strong electrolyte.

$AgCl$ is insoluble in water (per solubility rules). The net ionic equation is $Ag^+(aq) + Cl^-(aq) \to AgCl(s)$.

\(2H_2 + O_2 \to 2H_2O\). Moles \(H_2 = 10.0/2.02 = 4.95\text{ mol}\); moles \(O_2 = 10.0/32.0 = 0.313\text{ mol}\). The ratio requires \(2\text{ mol } H_2\) per \(1\text{ mol } O_2\). Available \(H_2/O_2 = 4.95/0.313 = 15.8\), far exceeding the required ratio of \(2\). So \(O_2\) is limiting.

The net ionic equation is \(H^+(aq) + OH^-(aq) \to H_2O(l)\). \(Na^+\) and \(Cl^-\) appear on both sides unchanged, so they are spectator ions.

Moles $CaCO_3 = 5.00/100.09 = 0.0500\text{ mol}$. The \(1:1\) ratio gives \(0.0500\text{ mol}\) $CaO$. Mass \(= 0.0500 \times 56.08 = 2.80\text{ g}\).

Oxidation is the loss of electrons. When a substance is oxidized, it loses electrons and its oxidation number increases. It acts as the reducing agent.

$NaOH + HCl \to NaCl + H_2O$ (\(1:1\) ratio). Moles $NaOH = 0.0250\text{ L} \times 0.100\text{ M} = 0.00250\text{ mol}$ = moles $HCl$. Molarity $HCl = 0.00250/0.0500 = 0.0500\text{ M}$.

Assume 100 g: 40.0 g C = 3.33 mol, 6.7 g H = 6.64 mol, 53.3 g O = 3.33 mol. Ratio C:H:O = 1:2:1, giving empirical formula CH2O.

$HNO_3$ is the oxidizing agent because nitrogen is reduced from \(+5\) in $HNO_3$ to \(+2\) in \(NO\). The oxidizing agent is the species that gets reduced.

Mass of water = 2.50 - 1.60 = 0.90 g. Moles CuSO4 = 1.60/159.6 = 0.01003 mol. Moles H2O = 0.90/18.02 = 0.04994 mol. Ratio = 0.04994/0.01003 = 5. Formula: CuSO4 * 5H2O.

$Pb(NO_3)_2 + 2KI \to PbI_2 + 2KNO_3$. Moles \(Pb^{2+} = 0.0100\text{ mol}\). Moles \(I^- = 0.0100\text{ mol}\). Need \(2\) \(I^-\) per \(Pb^{2+}\), so \(I^-\) is limiting: \(0.0100/2 = 0.00500\text{ mol}\) $PbI_2$. Mass \(= 0.00500 \times 461.0 = 2.31\text{ g}\).

Percent yield = (actual yield / theoretical yield) x 100 = (0.850/1.00) x 100 = 85.0%.

In a synthesis reaction, two or more reactants combine to form one product (A + B → AB). Choice B describes decomposition, choice C describes single displacement, and choice D describes double displacement (metathesis). All four reaction types are fundamental to AP Chemistry Unit 4.

Spectator ions appear unchanged on both sides of the complete ionic equation and are removed when writing the net ionic equation because they do not participate in the actual chemical change. Choices B, C, and D all describe ions that do participate in the reaction and would therefore appear in the net ionic equation.

Decomposition involves a single compound breaking into two or more simpler substances. Hydrogen peroxide breaking into water and oxygen gas fits this pattern perfectly. Choice B is synthesis (two elements combining), choice C is single displacement, and choice D is double displacement (precipitation).

Molar mass of CaCO3 = Ca + C + 3O = 40.08 + 12.01 + 3(16.00) = 40.08 + 12.01 + 48.00 = 100.09 g/mol. Choice D (56.08 g/mol) is the molar mass of CaO, a common distractor because CaCO3 decomposes to CaO and CO2. Choice B omits one oxygen atom.

Single displacement reactions follow the pattern A + BC → AC + B, where element A displaces element B from the compound. Choice B describes double displacement (metathesis). Choice C describes decomposition. Choice D describes synthesis. These distinctions are essential for predicting reaction products.

First balance C and H: C3H8 produces 3CO2 and 4H2O. Then count oxygen atoms on the right: 3(2) + 4(1) = 6 + 4 = 10 oxygen atoms needed, so coefficient for O2 = 10/2 = 5. The fully balanced equation is C3H8 + 5O2 → 3CO2 + 4H2O. Choice D (8) is the number of hydrogen atoms, a common error when students confuse atom counts with coefficients.

Acetic acid (CH3COOH) only partially ionizes in water, making it a weak acid (Ka = 1.8 x 10^-5). HCl, HNO3, and H2SO4 are all strong acids that dissociate essentially completely in aqueous solution. Knowing the six common strong acids (HCl, HBr, HI, HNO3, HClO4, H2SO4) is critical for writing net ionic equations correctly.

To react completely, 4.00 mol N2 would require 4.00 x 3 = 12.00 mol H2, but only 9.00 mol H2 is available, so H2 is the limiting reagent. Using H2: 9.00 mol H2 x (2 mol NH3 / 3 mol H2) = 6.00 mol NH3. Choice B incorrectly identifies N2 as limiting; if N2 were limiting, 4.00 x 2 = 8.00 mol NH3 would form, but that calculation is invalid here.

BaSO4 is insoluble (barium sulfate is an exception to the general solubility of sulfates). Na+ and Cl- are spectator ions and are removed from the net ionic equation. Choice B is the complete molecular equation, not the net ionic equation. Choice C incorrectly retains Cl- on both sides instead of canceling it.

Percent yield = (actual yield / theoretical yield) x 100 = (4.25 g / 5.00 g) x 100 = 85.0%. Choice B (117.6%) results from inverting the fraction (dividing theoretical by actual), which is a common algebraic error. Percent yield can never exceed 100% in a real experiment.

In KMnO4, K = +1 and each O = -2. The sum of all oxidation states must equal the overall charge of zero: (+1) + Mn + 4(-2) = 0, so Mn = +7. Choice B (+6) is the oxidation state of Mn in manganate ion (MnO42-). Choice C (+4) is found in MnO2. Recognizing that permanganate contains Mn(VII) is important for balancing redox reactions.

Using the dilution equation M1V1 = M2V2: (6.00 M)(V1) = (0.500 M)(250.0 mL), so V1 = (0.500 x 250.0) / 6.00 = 20.83 mL ≈ 20.8 mL. Choice D (125 mL) is the result of multiplying rather than dividing, a common procedural error. In lab practice, this concentrated acid is added to water, never the reverse.

Both HCl and NaOH are strong electrolytes that fully dissociate. Na+ and Cl- are spectator ions and cancel out. Only H+ and OH- participate in the chemical change, forming water. Choice B is the complete molecular equation. Choice D is the complete ionic equation before removing spectator ions. This net ionic equation is universal for all strong acid-strong base neutralizations.

Molar mass of C3H8 = 3(12.01) + 8(1.008) = 44.09 g/mol. Moles C3H8 = 22.0 / 44.09 = 0.499 mol. By stoichiometry, moles CO2 = 0.499 x 3 = 1.497 mol. Mass CO2 = 1.497 x 44.01 = 65.9 ≈ 66.0 g. Choice B (44.0 g) results from using a 1:1 mole ratio, ignoring that 3 mol CO2 forms per mol C3H8.

Most iodides are soluble, but PbI2 is a notable exception and forms a bright yellow precipitate. All nitrates are soluble, so KNO3 and Pb(NO3)2 remain in solution. Choice B incorrectly generalizes that all iodides are soluble — the exceptions include PbI2, AgI, and HgI2. Knowing the solubility rule exceptions is essential for predicting precipitation reactions.

Zn goes from an oxidation state of 0 to +2 (loses 2 electrons), so Zn is oxidized. Cu2+ goes from +2 to 0 (gains 2 electrons), so Cu2+ is reduced. SO42- is a spectator ion with no change in oxidation state. Choice B reverses the roles — a species cannot be reduced while losing electrons. Zn is the reducing agent; Cu2+ is the oxidizing agent.

The empirical formula mass of CH2O = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol. Dividing the molar mass by the empirical mass: 180 / 30.03 ≈ 6. Multiplying the empirical formula by 6 gives C6H12O6, which is glucose. Choice B gives a molar mass of 90 g/mol (factor of 3), and choice C gives 60 g/mol (factor of 2). Both are real molecules (lactic acid and acetic acid), illustrating that different molecular formulas can share the same empirical formula.

A single displacement reaction proceeds only when the free metal is more active (higher in the activity series) than the dissolved metal ion. Copper (Cu) is less active than zinc (Zn), so Cu cannot displace Zn2+ from solution. In choice B, Zn is more active than Cu, so the reaction proceeds. In C, Mg is more active than H, so H2 is released. In D, Fe is more active than Cu, so the reaction occurs.

Molar mass of Fe2O3 = 2(55.85) + 3(16.00) = 159.70 g/mol. Moles Fe2O3 = 1000 / 159.70 = 6.26 mol. Theoretical moles Fe = 6.26 x 2 = 12.52 mol. Theoretical mass Fe = 12.52 x 55.85 = 699 g. Actual yield = 699 x 0.875 = 612 g. Choice B (699 g) represents 100% theoretical yield — the most common error is forgetting to apply the percent yield to the calculated theoretical amount.

Cl2 is a stronger oxidizing agent than Br2, so it oxidizes Br- to Br2 while being reduced to Cl-. K+ is a spectator ion and is omitted from the net ionic equation. Choice C represents the reverse reaction, which does not occur spontaneously because Br2 is a weaker oxidizing agent than Cl2 and cannot oxidize Cl-. This reaction is used industrially to produce bromine.

Moles AgCl = 1.067 / 143.32 = 0.007446 mol total. Both salts produce 1 mol AgCl per mol, so mol NaCl + mol KCl = 0.007446. Combined with the mass equation: 58.44x + 74.55y = 0.500 g. Solving: y (KCl) = 0.004030 mol, x (NaCl) = 0.003416 mol. Mass NaCl = 0.003416 x 58.44 = 0.200 g. Percent = 0.200 / 0.500 x 100 = 40.0%. Choice B (60%) is the mass percent of KCl — a common error from solving for the wrong variable.

H2SO4 is diprotic: H2SO4 + 2NaOH → Na2SO4 + 2H2O. Moles H2SO4 = 0.0250 L x 0.200 mol/L = 0.00500 mol. Moles NaOH required = 2 x 0.00500 = 0.01000 mol. Volume NaOH = 0.01000 mol / 0.150 mol/L = 0.0667 L = 66.7 mL. Choice B (33.3 mL) is the most common error — treating H2SO4 as monoprotic and using a 1:1 mole ratio instead of 1:2.

From CO2: mol C = 2.000/44.01 = 0.04544 mol; mass C = 0.04544 x 12.01 = 0.546 g. From H2O: mol H = 2 x (0.818/18.02) = 0.09079 mol; mass H = 0.0915 g. Mass O = 1.000 - 0.546 - 0.0915 = 0.363 g; mol O = 0.363/16.00 = 0.02269 mol. Dividing by smallest (0.02269): C = 2.00, H = 4.00, O = 1.00. Empirical formula = C2H4O. Choice B (CH2O) represents a 1:2:1 ratio that does not match the calculated 2:4:1 ratio.

In excess ammonia, Cu2+ forms the tetraamminecopper(II) complex ion [Cu(NH3)4]2+, which has a characteristic deep blue color. NH3 acts as a ligand, donating electron pairs to Cu2+. Choice B describes only the initial precipitation of Cu(OH)2 that occurs with limited base. Choice D describes the partial reaction that forms the precipitate before excess NH3 dissolves it. This complex ion formation is an important example of how excess ligand can dissolve a precipitate.

Moles AlCl3 = 0.100 L x 0.300 M = 0.0300 mol. Moles NaOH = 0.150 L x 0.200 M = 0.0300 mol. AlCl3 requires 3 mol NaOH per mol, so 0.0300 mol AlCl3 needs 0.0900 mol NaOH — far more than the 0.0300 mol available. NaOH is the limiting reagent. Moles Al(OH)3 = 0.0300 mol NaOH x (1/3) = 0.0100 mol. Mass = 0.0100 x 78.00 = 0.780 g. Choice B (2.34 g) is the most common error — assuming AlCl3 is the limiting reagent (0.0300 mol x 78.00 = 2.34 g) while ignoring the 3:1 stoichiometry of NaOH.

A synthesis reaction involves two or more reactants combining to form a single product. In choice B, sodium metal and chlorine gas combine to form sodium chloride — one product from two reactants. Choice A is a decomposition reaction (one reactant breaks into two products), choice C is a single-replacement reaction, and choice D is a double-replacement (neutralization) reaction.

The net ionic equation is Ag+ + Cl- → AgCl(s). Sodium (Na+) and nitrate (NO3-) ions remain dissolved and unchanged throughout the reaction — they are the spectator ions. Ag+ and Cl- actively participate by forming the insoluble precipitate and therefore do not appear in the net ionic equation.

The balanced equation shows a 2:2 (or 1:1) mole ratio between H2 and H2O. Therefore, 3.0 moles of H2 produces 3.0 moles of H2O. Choice A (1.5 mol) results from applying the 2:1 ratio of H2 to O2 rather than H2 to H2O. Choice D incorrectly doubles the mole count.

A decomposition reaction has a single reactant that breaks down into two or more products. KClO3 decomposes into KCl and O2 when heated. Choice A is a synthesis reaction (two reactants, one product), choice B is a single-replacement reaction, and choice D is a double-replacement (precipitation) reaction.

BaSO4 is insoluble in water and forms a white precipitate. Solubility rules state that sulfates of Ba2+, Pb2+, and Ca2+ are insoluble exceptions. NaNO3 is soluble because nearly all nitrates and sodium salts are soluble. KCl is soluble because most chlorides are soluble. (NH4)2SO4 is soluble because ammonium salts and most sulfates are soluble.

The molar mass of H2O = 2(1.0) + 16.0 = 18.0 g/mol. Percent O = (16.0 / 18.0) x 100 = 88.9%. Choice A (11.1%) is the mass percent of hydrogen, not oxygen. Choices B and C reflect errors in calculating the molar mass or using incorrect ratios.

In a single-replacement reaction, one element displaces another from a compound. Magnesium metal replaces hydrogen in HCl, producing MgCl2 and H2 gas. Choice A is a synthesis reaction, choice B is a decomposition reaction, and choice D is a double-replacement (metathesis) reaction.

The balanced equation shows a 3:2 mole ratio between H2 and NH3. For 6.0 mol H2: 6.0 x (2/3) = 4.0 mol NH3. Choice D (6.0 mol) incorrectly applies a 1:1 ratio. Choice B (3.0 mol) uses a 2:1 ratio in the wrong direction. Choice A (2.0 mol) results from applying the N2:NH3 ratio to H2 instead.

Convert grams of Al to moles: 54.0 g / 27.0 g/mol = 2.00 mol Al. The 4:2 mole ratio (Al to Al2O3) gives 2.00 x (2/4) = 1.00 mol Al2O3. Mass = 1.00 mol x 102.0 g/mol = 102.0 g. Choice A (51.0 g) results from a 4:1 ratio error. Choice B incorrectly applies the Al:O2 ratio (4:3) instead of the Al:Al2O3 ratio.

First find the theoretical yield: 4.86 g / 24.3 g/mol = 0.200 mol Mg. Since the Mg:MgCl2 mole ratio is 1:1, theoretical yield = 0.200 mol x 95.2 g/mol = 19.04 g. Percent yield = (17.7 g / 19.04 g) x 100 = 92.9%. Choice A results from incorrectly using 100 g/mol for MgCl2. Percent yield cannot exceed 100%, so any answer above the theoretical yield is impossible.

Acetic acid is a weak acid and remains in molecular form (CH3COOH) in the net ionic equation — it is not written as H+. Sodium hydroxide is a strong base that fully dissociates, so OH- appears as a free ion. Choice A applies only when a strong acid reacts with a strong base, where H+ exists as a free ion. Choice D is the complete ionic equation, not the net ionic equation, because Na+ appears on both sides as a spectator ion.

Track oxidation state changes for each sulfur species. In H2S, sulfur has an oxidation state of -2. In elemental sulfur (S), the oxidation state is 0. An increase in oxidation state represents oxidation (loss of electrons), so sulfur in H2S is oxidized from -2 to 0. The sulfur in SO2 goes from +4 to 0, which is a decrease — that is reduction (gain of electrons). Hydrogen remains +1 and oxygen remains -2 throughout.

First calculate moles of HCl: 0.1500 L x 0.400 mol/L = 0.0600 mol HCl. Since the HCl:NaCl mole ratio is 1:1, 0.0600 mol NaCl is produced. Mass = 0.0600 mol x 58.44 g/mol = 3.51 g. Choice A results from using 100 mL instead of 150.0 mL. Choice C doubles the moles incorrectly.

Assume 100 g of compound: 40.0 g S and 60.0 g O. Convert to moles: 40.0 / 32.1 = 1.25 mol S and 60.0 / 16.0 = 3.75 mol O. Divide by the smallest value (1.25): S = 1.00, O = 3.00. The empirical formula is SO3. Choice B (SO2) would require a 1:2 ratio. Choice D (S2O3) corresponds to approximately 46% S and 54% O by mass.

The balanced equation shows a 1:3 mole ratio between C3H8 and CO2. For 5.00 mol C3H8: 5.00 x 3 = 15.0 mol CO2. At STP, 1 mole of any ideal gas occupies 22.4 L, so volume = 15.0 mol x 22.4 L/mol = 336 L. Choice A (112 L) uses a 1:1 ratio. Choice B (224 L) uses a 1:2 ratio, which applies to CO2 from methane combustion, not propane.

The net ionic equation includes only species that participate in the reaction. Ba2+ and SO42- combine to form insoluble BaSO4(s). Na+ and Cl- are spectator ions and are omitted. Choice A is the complete molecular equation, not the net ionic equation. Choice C incorrectly suggests NaCl precipitates — NaCl is soluble, so Na+ and Cl- remain in solution as spectator ions.

First find moles of CO2: 8.80 g / 44.0 g/mol = 0.200 mol CO2. The balanced equation shows a 1:1 mole ratio between CaCO3 and CO2, so 0.200 mol CaCO3 is required. Mass of CaCO3 = 0.200 mol x 100.1 g/mol = 20.0 g. Choice B (10.0 g) would only produce 4.40 g of CO2. Choice A results from incorrectly using the ratio of molar masses directly without converting through moles.

Calculate moles of each reactant: Fe = 11.2 / 55.8 = 0.2007 mol; O2 = 6.40 / 32.0 = 0.200 mol. Determine how much Fe2O3 each can produce: 0.2007 mol Fe x (2/4) = 0.1003 mol Fe2O3; 0.200 mol O2 x (2/3) = 0.1333 mol Fe2O3. Iron produces less product, so Fe is the limiting reagent. Theoretical yield = 0.1003 mol x 159.7 g/mol = 16.0 g. Choice C incorrectly identifies O2 as the limiting reagent, though O2 can produce more product than Fe can.

Initial moles HCl = 0.0300 L x 0.500 mol/L = 0.0150 mol. Moles of excess HCl = moles NaOH used = 0.0100 L x 0.250 mol/L = 0.00250 mol. Moles HCl that reacted with CaCO3 = 0.0150 - 0.00250 = 0.01250 mol. Since CaCO3 requires 2 mol HCl per mole, moles CaCO3 = 0.01250 / 2 = 0.006250 mol. Mass CaCO3 = 0.006250 x 100.1 = 0.626 g. Percent purity = (0.626 / 1.000) x 100 = 62.6%. Choice A (31.3%) results from forgetting to divide by 2 in the stoichiometry step for the diprotic relationship.

Moles of glucose = 90.1 g / 180.2 g/mol = 0.500 mol. Using the 1:2 mole ratio, theoretical moles of ethanol = 0.500 x 2 = 1.00 mol. Theoretical mass = 1.00 mol x 46.07 g/mol = 46.07 g. Applying 85.0% yield: 46.07 x 0.850 = 39.2 g. Choice C (46.1 g) is the theoretical yield before applying percent yield. Choice D incorrectly doubles the theoretical yield.

Manganese in MnO4- has an oxidation state of +7. In the product Mn2+, it is +2. The decrease from +7 to +2 means each Mn gains 5 electrons (reduction). Iron goes from Fe2+ (+2) to Fe3+ (+3), losing 1 electron per ion (oxidation). The equation balances because 5 Fe2+ each donate 1 electron, totaling 5 electrons — exactly matching the 5 electrons gained by one MnO4-. Choice C confuses the +7 oxidation state of Mn with the number of electrons transferred.

Find moles of each element from combustion products. Moles C = moles CO2 = 1.532 / 44.01 = 0.03481 mol. Moles H = 2 x (0.732 / 18.02) = 0.08124 mol. The C:H mole ratio = 0.03481:0.08124; dividing by 0.03481 gives 1:2.33, which simplifies to 3:7. The empirical formula is C3H7, with mass = 3(12.01) + 7(1.008) = 43.1 g/mol. Since the molar mass is 86.2 g/mol, n = 86.2 / 43.1 = 2. The molecular formula is C6H14. Choice C (C6H12) has a 1:2 C:H ratio and is an alkene or cycloalkane with empirical formula CH2.

When a weak acid (NH4+) donates a proton to a weak base (CH3COO-), ammonia (NH3) and acetic acid (CH3COOH) are formed — both weak electrolytes that remain in molecular form. Na+ and Cl- are spectator ions and are excluded from the net ionic equation. Choice C (H+ + OH- → H2O) applies only when strong acids and strong bases react, producing free H+ and OH- ions. Choice D is the full molecular equation.

Step 1: 56.0 g N2 / 28.0 g/mol = 2.00 mol N2. Using the 1:2 mole ratio (N2:NH3), 2.00 mol N2 produces 4.00 mol NH3. Step 2: The NH3:NO mole ratio is 4:4 (1:1), so 4.00 mol NH3 produces 4.00 mol NO. Mass of NO = 4.00 mol x 30.0 g/mol = 120.0 g. Choice B (60.0 g) results from using only 1.00 mol N2 or incorrectly halving the moles when converting N2 to NH3.

Calculate NaOH needed for each acid separately. H2SO4 is diprotic and requires 2 mol NaOH per mole: 0.200 mol x 2 = 0.400 mol NaOH. HCl requires 1 mol NaOH per mole: 0.300 mol x 1 = 0.300 mol NaOH. Total NaOH = 0.400 + 0.300 = 0.700 mol. Volume = 0.700 mol / 2.00 mol/L = 0.350 L = 350 mL. Choice B (250 mL) results from incorrectly treating H2SO4 as monoprotic, yielding only 0.500 mol total NaOH needed.

A synthesis reaction involves two or more substances combining to form a single product. 2Na + Cl2 → 2NaCl shows two reactants combining into one product. Choice B is a decomposition reaction (one reactant splits into two products), choice C is a single-displacement reaction, and choice D is a double-displacement reaction.

A decomposition reaction has one reactant breaking down into two or more products. 2H2O2 → 2H2O + O2 shows hydrogen peroxide decomposing into water and oxygen gas. Choice A is a synthesis reaction, choice B is a single-displacement reaction, and choice D is a double-displacement reaction.

Al2(SO4)3 contains 2 Al atoms, 3 S atoms, and 12 O atoms (4 oxygens per sulfate group times 3 groups). Molar mass = 2(26.98) + 3(32.07) + 12(16.00) = 53.96 + 96.21 + 192.00 = 342.17 g/mol. A common error is counting only 4 total oxygen atoms rather than 12, which gives choice C.

This is a single displacement (also called single replacement) reaction because zinc metal displaces hydrogen from hydrochloric acid. One element replaces another element in a compound. Double displacement involves two compounds exchanging ions, decomposition produces multiple products from one reactant, and synthesis combines reactants into one product.

Spectator ions appear unchanged on both sides of the ionic equation and do not participate in the actual reaction. Since Ag+ and Cl- combine to form the precipitate AgCl, they are the reactive ions. Na+ and NO3- remain dissolved in solution throughout the reaction and are the spectator ions. The net ionic equation is simply Ag+(aq) + Cl-(aq) → AgCl(s).

Each formula unit of H2SO4 contains 4 oxygen atoms. Therefore, 2.0 mol of H2SO4 contains 2.0 × 4 = 8.0 mol of oxygen atoms. A common mistake is to count only 1 oxygen per formula unit (choice A) or to confuse oxygen atoms with oxygen molecules O2 and give 4.0 mol (choice B).

Combustion reactions involve a fuel reacting with oxygen gas, producing carbon dioxide and water for hydrocarbons. C3H8 + 5O2 → 3CO2 + 4H2O shows propane burning in oxygen. Choice A is a decomposition reaction, choice B is a double-displacement acid-base reaction that produces a precipitate, and choice C is a single-displacement reaction between sodium and water.

Moles of N2 = 14.0 g / 28.02 g/mol = 0.4997 mol. The molar ratio is 1 mol N2 : 2 mol NH3, so moles of NH3 = 0.4997 × 2 = 0.9994 mol. Mass of NH3 = 0.9994 mol × 17.03 g/mol = 17.0 g. Choice A (8.51 g) results from using a 1:1 ratio instead of 1:2. Choice D (34.1 g) doubles the correct answer due to a ratio inversion error.

Since HCl and NaOH are both strong electrolytes, they fully dissociate in solution. Na+ and Cl- are spectator ions that do not change. The only reaction occurring is H+(aq) + OH-(aq) → H2O(l). Choice A is the complete molecular equation rather than the net ionic equation. Choice C is incorrect because NaCl remains dissolved and does not precipitate from solution.

In NH4+, the overall charge is +1. Hydrogen bonded to a nonmetal is assigned +1. Setting up the equation: N + 4(+1) = +1, so N = +1 - 4 = -3. This is the same oxidation state as nitrogen in NH3. A common confusion is with NO3-, where nitrogen is +5, but in NH4+ the nitrogen is in its most reduced state, surrounded by hydrogen atoms.

Assuming 100 g: C = 52.17 g / 12.01 g/mol = 4.344 mol; H = 13.04 g / 1.008 g/mol = 12.94 mol; O = 34.78 g / 16.00 g/mol = 2.174 mol. Divide by the smallest value (2.174): C = 2.0, H = 5.95 ≈ 6, O = 1.0. Empirical formula = C2H6O. Choice A (CH2O) corresponds to formaldehyde and would require approximately 40% C, 6.7% H, and 53.3% O, which does not match the given percentages.

Moles of CaCO3 = 10.0 g / 100.09 g/mol = 0.09991 mol. From the balanced equation, 1 mol CaCO3 requires 2 mol HCl, so moles of HCl needed = 0.09991 × 2 = 0.1998 mol. Volume = moles / molarity = 0.1998 mol / 0.500 mol/L = 0.3996 L = 400 mL. Choice B (200 mL) is a common error from using a 1:1 ratio and ignoring the coefficient of 2 for HCl in the balanced equation.

Lead(II) iodide has the formula PbI2, requiring one Pb2+ and two I- ions to balance the charge. The net ionic equation removes spectator ions (K+ and NO3-), leaving Pb2+(aq) + 2I-(aq) → PbI2(s). Choice B is incorrect because a 1:1 ratio gives PbI, which is not a stable compound — lead(II) has a 2+ charge that requires two iodide ions. Choice C is the complete molecular equation, not the net ionic equation.

Moles of Al = 5.40 g / 26.98 g/mol = 0.2001 mol. The molar ratio of Al to Cu is 2:3, so moles of Cu = 0.2001 × (3/2) = 0.3002 mol. Mass of Cu = 0.3002 mol × 63.55 g/mol = 19.1 g. Choice B (12.7 g) results from incorrectly using a 1:1 molar ratio. Choice A (6.37 g) results from using a 2:1 ratio, inverting the actual stoichiometry.

The oxidizing agent is the species that is reduced (gains electrons). In this reaction, each oxygen atom in O2 goes from oxidation state 0 to -2 in MgO, meaning O2 gains electrons and is reduced — making O2 the oxidizing agent. Mg goes from 0 to +2 (loses electrons), so Mg is the reducing agent. MgO is a product and cannot serve as either an oxidizing or reducing agent in this reaction.

Percent yield = (actual yield / theoretical yield) × 100 = (18.5 g / 22.0 g) × 100 = 84.1%. Choices B and C result from arithmetic errors. Choice D (119%) is physically impossible — the actual yield can never exceed the theoretical yield without violating conservation of mass. Any percent yield above 100% indicates an error in measurement or calculation.

CaCl2 and Na2SO4 react to form CaSO4, which is sparingly soluble and precipitates: Ca2+(aq) + SO42-(aq) → CaSO4(s). Choice A produces NaCl and KNO3, both soluble — no precipitate. Choice B is an acid-base neutralization producing soluble NaCl and water. Choice C produces KCl and NaOH, both of which remain dissolved in solution.

This is a double displacement (metathesis) reaction where the cation of one compound pairs with the anion of the other: FeCl3 + 3NaOH → Fe(OH)3(s) + 3NaCl. Iron(III) hydroxide, Fe(OH)3, is an insoluble rust-colored solid that precipitates. Choice A is incorrect because no free metal forms — iron remains as a cation in the precipitate. Single displacement would require a metal to displace another ion from solution.

Reduction half-reaction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O (each Cr goes from +6 to +3, gaining 3e- per Cr atom, so 6e- total for two Cr). Oxidation half-reaction: Fe2+ → Fe3+ + e- (multiplied by 6 to balance electrons: 6Fe2+ → 6Fe3+ + 6e-). Adding both: Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O. Choice B uses an incorrect 3:1 Fe:Cr2O7 ratio — 6 electrons must be transferred, requiring 6 Fe2+ ions.

Moles of CO2 = 0.336 L / 22.4 L/mol = 0.0150 mol. From the balanced equation, mol Na2CO3 = mol CO2 = 0.0150 mol (1:1 ratio). Mass of Na2CO3 = 0.0150 mol × 105.99 g/mol = 1.590 g. Mass percent = (1.590 g / 2.00 g) × 100 = 79.5%. NaCl does not react with HCl to produce any gas, so all CO2 comes exclusively from Na2CO3. Choice A likely results from using an incorrect 2:1 CO2:Na2CO3 ratio.

Theoretical yield: mol H2O2 = 6.80 / 34.02 = 0.1999 mol. The molar ratio is 2 mol H2O2 : 1 mol O2, so theoretical mol O2 = 0.1999 / 2 = 0.09995 mol. Theoretical mass O2 = 0.09995 × 32.00 = 3.20 g. Percent yield = (1.44 / 3.20) × 100 = 45.0%. A common error is using a 1:1 molar ratio, which gives a theoretical yield of 6.40 g and an erroneously low percent yield near 22%.

Reduction: MnO4- + 2H2O + 3e- → MnO2 + 4OH- (multiply by 2: 2MnO4- + 4H2O + 6e- → 2MnO2 + 8OH-). Oxidation: SO32- + 2OH- → SO42- + H2O + 2e- (multiply by 3: 3SO32- + 6OH- → 3SO42- + 3H2O + 6e-). Adding and simplifying (cancel 3H2O and 6OH-): 2MnO4- + 3SO32- + H2O → 2MnO2 + 3SO42- + 2OH-. Choice B omits the water and hydroxide, violating atom and charge balance. Choice D shows Mn reduced to Mn2+, which occurs in acidic conditions, not basic.

Moles of SO2 = 160.0 / 64.07 = 2.497 mol. Step 1: theoretical SO3 = 2.497 mol (1:1 ratio); actual SO3 = 2.497 × 0.900 = 2.247 mol. Step 2: theoretical H2SO4 = 2.247 mol (1:1 ratio); actual H2SO4 = 2.247 × 0.950 = 2.135 mol. Mass = 2.135 × 98.09 = 209 g. Choice D (247 g) assumes 100% yield for both steps. A common error in multi-step problems is applying only one of the two percent yields.

Moles of NaOH used = 0.03250 L × 0.1200 mol/L = 3.900 × 10-3 mol. The reaction HCl + NaOH → NaCl + H2O has a 1:1 molar ratio, so moles of HCl = 3.900 × 10-3 mol. Molarity of HCl = (3.900 × 10-3 mol) / (0.02500 L) = 0.1560 M. Choice A results from dividing the NaOH molarity by the volume ratio rather than multiplying. At the equivalence point, moles of acid always equal moles of base for a monoprotic system.

Moles of Cu = 19.05 / 63.55 = 0.2998 mol; moles of HNO3 = 0.0400 L × 8.00 mol/L = 0.3200 mol. To consume all Cu: 0.2998 × (8/3) = 0.7995 mol HNO3 needed, but only 0.3200 mol is available — HNO3 is the limiting reagent. Moles of NO produced from HNO3: 0.3200 × (2/8) = 0.08000 mol. Mass of NO = 0.08000 × 30.01 = 2.40 g. A common error is assuming Cu is limiting simply because its mass is large, without comparing moles to the required stoichiometric ratio.

A decomposition reaction involves a single reactant breaking down into two or more products. In choice B, hydrogen peroxide (one compound) decomposes into water and oxygen gas. Choice A is a synthesis reaction (two reactants combine into one product), choice C is a single-displacement reaction, and choice D is a double-displacement (precipitation) reaction.

The reaction produces insoluble AgCl as a precipitate, so Ag+ and Cl- participate in the reaction and are NOT spectator ions. Na+ and NO3- remain fully dissolved and unchanged — they appear on both sides of the complete ionic equation and are therefore the spectator ions that are eliminated when writing the net ionic equation.

Complete combustion of any hydrocarbon in excess oxygen yields only carbon dioxide and water. The balanced equation is C3H8 + 5O2 → 3CO2 + 4H2O. Carbon monoxide (choice A) is a product of incomplete combustion, which occurs when oxygen supply is limited. Molecular hydrogen (choice D) is never a product of hydrocarbon combustion.

Coefficients in a balanced equation directly give mole ratios. The coefficient of H2 is 3 and the coefficient of NH3 is 2, so H2:NH3 = 3:2. For every 2 moles of ammonia produced, exactly 3 moles of hydrogen are consumed. Choice C (2:3) inverts the ratio, and choice D (1:2) confuses the N2:NH3 ratio with the H2:NH3 ratio.

A synthesis reaction has the general form A + B → AB: two or more reactants combine to form a single product. Choice B fits this pattern — two elements (sodium and chlorine) combine into one compound (sodium chloride). Choice A is a decomposition reaction, choice C is a single-displacement reaction, and choice D is a double-displacement (precipitation) reaction.

The net ionic equation retains only the ions that directly participate in forming the precipitate BaSO4, eliminating spectator ions Na+ and Cl-. Choice B is the molecular (formula unit) equation. Choice C is the complete ionic equation — it still includes spectator ions on both sides. Choice D incorrectly shows only the spectator ions reacting.

The limiting reagent is the reactant consumed entirely, stopping further reaction and capping the theoretical yield. It is not necessarily the reactant with the smallest mass (A) or fewest moles (B) — these must be evaluated relative to stoichiometric ratios. For example, 1 mole of a reactant can be in excess over 10 moles of another if the mole ratio demands it. Molar mass (D) is unrelated to determining the limiting reagent.

Both HCl (strong acid) and NaOH (strong base) dissociate completely; the product NaCl is also fully dissociated. Removing the spectator ions Na+ and Cl- from the complete ionic equation (choice C) leaves only H+ combining with OH- to form water. This same net ionic equation applies to all strong acid-strong base neutralizations, regardless of the specific ions involved. Choice A is the molecular equation.

To identify the limiting reagent, determine how much H2 is needed for 4.00 mol N2: 4.00 mol N2 x (3 mol H2 / 1 mol N2) = 12.0 mol H2. Since only 9.00 mol H2 is available, H2 is the limiting reagent. Moles of NH3 from H2: 9.00 mol H2 x (2 mol NH3 / 3 mol H2) = 6.00 mol NH3. Choice A incorrectly identifies N2 as limiting and computes yield based on N2 alone.

The blue precipitate is Cu(OH)2. Cu2+ and OH- are the ions that combine to form it; Na+ and SO42- remain dissolved and are spectator ions. Choice C is the correct molecular equation but not the net ionic equation. Choice A is wrong because CuSO4 is a soluble salt, not a precipitate — Cu2+ and SO42- do not combine to precipitate.

Molar mass of NH4NO3 = 14.01 + 4(1.008) + 14.01 + 3(16.00) = 80.05 g/mol. NH4NO3 contains two nitrogen atoms, so total mass of N = 2 x 14.01 = 28.02 g/mol. Percent N = (28.02 / 80.05) x 100 = 35.00%. Choice A (17.50%) results from counting only one nitrogen atom. Choice C (28.02%) mistakenly reports the raw mass of both nitrogen atoms as if it were already a percentage.

The empirical formula mass of CH2O = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol. Dividing the molar mass by the empirical formula mass: 180 / 30.03 = 6. Multiplying each subscript by 6 gives C6H12O6, the molecular formula of glucose. Choices A, B, and C correspond to multipliers of 2, 3, and 4 respectively — none of these yield a molar mass near 180 g/mol.

Moles of Na2CO3 = 0.2500 g / 105.99 g/mol = 0.002359 mol. By the 1:2 stoichiometric ratio, moles of HCl = 2 x 0.002359 = 0.004718 mol. Molarity = 0.004718 mol / 0.02350 L = 0.2008 M. Choice A (0.1004 M) results from forgetting to double the moles of Na2CO3 to account for the 2:1 HCl:Na2CO3 ratio — a common stoichiometry error in acid-base titrations.

Moles of H2 = 8.00 g / 2.016 g/mol = 3.968 mol. The mole ratio of H2O to H2 from the balanced equation is 2:2 = 1:1, so moles of H2O = 3.968 mol. Mass of H2O = 3.968 mol x 18.02 g/mol = 71.5 g. Choice A (18.0 g) is the mass of exactly 1 mole of water. Choice B (36.0 g) results from using 2.0 g/mol (approximately) for H2 and losing track of the 1:1 ratio.

A single-displacement reaction occurs only when the free metal is more active (higher in the activity series) than the metal it would displace. In choice C, silver (Ag) would need to displace copper (Cu), but copper is more active than silver — this reaction does not occur. Choice A occurs because Zn is above H in the activity series. Choice B occurs because Cu is above Ag. Choice D occurs because Fe is above Cu.

Because acetic acid is a weak acid, it is written in molecular (undissociated) form in ionic equations. NaOH is a strong base, so OH- appears as a free ion; Na+ is a spectator ion that is removed. The net ionic equation retains HC2H3O2 as a molecule, which distinguishes it from the strong acid case. Choice A applies only when a strong acid reacts with a strong base. Choice D is the molecular equation. Choice C still includes the spectator ion Na+.

Assign known oxidation states: K = +1 and O = -2. Setting up the equation for the overall neutral compound: 2(+1) + 2(Cr) + 7(-2) = 0, which gives 2 + 2Cr - 14 = 0, so Cr = +6. The +6 state is characteristic of dichromate and chromate ions, making them powerful oxidizing agents. Choice A (+3) is the oxidation state of Cr in Cr2O3 and in Cr3+ (aq) — the reduced product after dichromate acts as an oxidant.

Moles of Fe2O3 = 160.0 / 159.7 = 1.002 mol; moles of CO = 84.0 / 28.01 = 2.999 mol. The stoichiometric ratio requires 3 mol CO per mol Fe2O3, so 1.002 mol Fe2O3 needs 3.006 mol CO. Since only 2.999 mol CO is available, CO is the limiting reagent. Moles of Fe from CO = 2.999 x (2/3) = 1.999 mol. Mass of Fe = 1.999 x 55.85 = 111.6 g. Choices C and D are larger than the mass of Fe2O3 used, which is physically impossible.

Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O (5 electrons gained per Mn). Oxidation half-reaction: H2C2O4 → 2CO2 + 2H+ + 2e- (carbon goes from +3 to +4; 2 electrons lost per molecule). To equalize electrons, multiply the reduction by 2 and the oxidation by 5 (LCM = 10). This yields 5H2C2O4. The balanced equation is 2MnO4- + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O. Choice D (8) is the coefficient of water, a common mix-up.

Moles of Al = 10.8 / 26.98 = 0.4003 mol; moles of Cl2 = 21.3 / 70.90 = 0.3004 mol. For the 2:3 Al:Cl2 ratio, 0.4003 mol Al requires 0.4003 x (3/2) = 0.6005 mol Cl2. Only 0.3004 mol Cl2 is available, so Cl2 is the limiting reagent. Moles of AlCl3 = 0.3004 x (2/3) = 0.2003 mol. Mass = 0.2003 x 133.3 = 26.7 g. Choice C (53.4 g) results from incorrectly treating Al as the limiting reagent.

Total moles HCl added = 0.05000 L x 0.2000 M = 0.01000 mol. Moles of excess HCl neutralized by NaOH = 0.01000 L x 0.1000 M = 0.001000 mol. HCl that reacted with CaCO3 = 0.01000 - 0.001000 = 0.009000 mol. Using the 1:2 CaCO3:HCl ratio, moles CaCO3 = 0.009000 / 2 = 0.004500 mol. Mass of pure CaCO3 = 0.004500 x 100.09 = 0.4504 g. Percent purity = (0.4504 / 0.6000) x 100 = 75.1%.

Since HC2H3O2 is a weak acid, it is written in molecular form — not as free H+. Since CaCO3 is a solid, it is not broken into ions. The soluble products Ca2+ and C2H3O2- are written in ionic form. With no spectator ions to cancel, the net ionic equation equals the full ionic equation as shown in choice B. Choice A incorrectly treats acetic acid as a strong acid. Choice C incorrectly shows solid CaCO3 as dissolved CO32-(aq).

Moles of CO2 = 0.7333 / 44.01 = 0.01666 mol, so moles of C = 0.01666 mol (mass C = 0.2001 g). Moles of H2O = 0.3000 / 18.02 = 0.01665 mol, so moles of H = 2 x 0.01665 = 0.03330 mol (mass H = 0.03357 g). Mass of O = 0.5000 - 0.2001 - 0.03357 = 0.2663 g; moles of O = 0.2663 / 16.00 = 0.01664 mol. Dividing by the smallest value gives C:H:O = 1:2:1, so the empirical formula is CH2O. Choices A and D require different H:O ratios than observed.

Moles of NaHCO3 = 21.0 / 84.01 = 0.2500 mol. From the 2:1 stoichiometry in the balanced equation, moles of Na2CO3 = 0.2500 / 2 = 0.1250 mol. Molarity = 0.1250 mol / 0.2500 L = 0.500 M. Choice A (0.250 M) results from halving again unnecessarily. Choice C (1.00 M) results from neglecting the 2:1 ratio and using 0.2500 mol Na2CO3 directly.

Reduction half-reaction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O (balance O with H2O, then H with H+, then charge with electrons). Oxidation half-reaction: Sn2+ → Sn4+ + 2e-. To equalize electrons (LCM of 6 and 2 is 6), multiply the oxidation half-reaction by 3: 3Sn2+ → 3Sn4+ + 6e-. Adding the half-reactions: Cr2O72- + 3Sn2+ + 14H+ → 2Cr3+ + 3Sn4+ + 7H2O. The coefficient of H+ is 14. Choice A (7) equals the number of water molecules, a frequent confusion when reading the balanced equation.

A synthesis reaction combines two or more reactants to form a single product. In choice B, sodium metal and chlorine gas combine to form one product, sodium chloride. Choice A is a decomposition reaction (one reactant breaks into two products). Choice C is a single displacement reaction. Choice D is a double displacement (precipitation) reaction.

The law of conservation of mass states that the total mass of reactants equals the total mass of products. Total reactant mass = 25.0 + 15.0 = 40.0 g. Since 30.0 g of product C is formed, product D must account for the remaining mass: 40.0 - 30.0 = 10.0 g. Choice A (5.0 g) incorrectly subtracts only one reactant mass. Choice D (40.0 g) represents the total reactant mass, not the mass of D alone.

BaSO4 is insoluble because sulfates of barium, lead(II), and calcium are insoluble exceptions to the general rule that sulfates are soluble. NaNO3 is soluble because all nitrates are soluble and all sodium salts are soluble. KCl is soluble because all potassium salts and all chlorides (except those of Ag+, Pb2+, and Hg2^2+) are soluble. NH4Cl is soluble because all ammonium salts are soluble.

Moles = mass / molar mass = 44.0 g / 44.01 g/mol = 1.00 mol. Choice A (0.500 mol) would correspond to 22.0 g. Choice C (2.00 mol) would require 88.0 g. Choice D confuses grams with moles by treating the numerical value of the mass as the number of moles.

Using oxidation state rules: each H is +1 and each O is -2. Setting up the equation: 2(+1) + S + 4(-2) = 0, which gives 2 + S - 8 = 0, so S = +6. Choice B (+4) is the oxidation state of sulfur in SO2 and H2SO3. Choice A (+2) and choice D (+8) do not correspond to common sulfur compounds; +8 exceeds the maximum possible for sulfur.

A decomposition reaction has one reactant breaking down into two or more products. In choice B, one reactant (H2O2) produces two products (H2O and O2). Choice A is a synthesis reaction (two reactants form one product). Choice C is also a synthesis reaction. Choice D is a double displacement (metathesis) reaction where two compounds exchange ions.

Mass = moles x molar mass = 2.00 mol x 58.44 g/mol = 116.88 g, which rounds to 116.9 g. Choice A (29.2 g) would be the mass of 0.500 mol. Choice B (58.4 g) represents only 1.00 mol, a common error of forgetting to multiply by the number of moles. Choice D (175.3 g) corresponds to 3.00 mol.

Percent yield = (actual yield / theoretical yield) x 100 = (7.23 g / 8.50 g) x 100 = 85.1%. Choice B (117.6%) results from inverting the ratio (theoretical / actual x 100), which would imply more product was made than theoretically possible. Choice C (14.9%) is the percent loss, not the percent yield. Choice D (74.5%) comes from an arithmetic error.

To find the limiting reagent, compare mole ratios. Mol N2 = 14.0 / 28.02 = 0.4996 mol; mol H2 = 3.0 / 2.02 = 1.485 mol. To consume all the N2, the reaction requires 3 x 0.4996 = 1.499 mol H2, but only 1.485 mol is available, so H2 is limiting. Choice A is wrong because having fewer moles does not determine the limiting reagent; the stoichiometric ratio must be considered. Choice D is wrong because molar mass is irrelevant to limiting reagent identification.

Lead(II) iodide (PbI2) is insoluble and forms as a yellow precipitate. The net ionic equation shows only the ions that participate in the reaction. Since Pb2+ forms a 2+ cation and iodide is I-, two iodide ions are needed to balance the charge: Pb2+(aq) + 2I-(aq) → PbI2(s). Choice B incorrectly uses a 1:1 ratio and writes an impossible formula PbI. Choice C is the complete molecular equation, not the net ionic equation. Choice D incorrectly shows spectator ions recombining.

First calculate moles of NaOH: 10.0 g / 40.00 g/mol = 0.250 mol. Then divide by volume in liters: M = 0.250 mol / 0.5000 L = 0.500 M. Choice A (0.200 M) results from dividing 0.250 mol by 1.25 L (an arithmetic error). Choice D (2.00 M) results from forgetting to convert mL to L, dividing by 0.500 instead of multiplying. Choice B (0.400 M) uses an incorrect calculation.

Using the dilution equation M1V1 = M2V2: (6.0 M)(250.0 mL) = (1.5 M)(V2), so V2 = 1500 / 1.5 = 1000.0 mL. This is the final total volume. The volume of water to add is 1000.0 - 250.0 = 750.0 mL. Choice D (1000.0 mL) is the final total volume, not the volume of water added — a common error. Choice A is simply the original volume, which is incorrect.

This is a single displacement reaction: one element (Cl2) displaces another element (Br-) from a compound. Bromine is oxidized: Br- goes from an oxidation state of -1 to 0 in Br2, losing electrons. Chlorine is reduced: Cl2 goes from 0 to -1. Choice C incorrectly identifies chlorine as oxidized; chlorine gains electrons (is reduced). Choice D correctly identifies chlorine as reduced but misclassifies the reaction type as double displacement.

Using the mole ratio from the balanced equation: 2 mol C8H18 requires 25 mol O2. For 4.00 mol C8H18: 4.00 mol C8H18 x (25 mol O2 / 2 mol C8H18) = 50.0 mol O2. Choice B (25.0 mol) uses the coefficient from the balanced equation directly without applying the correct mole ratio — it treats 2 mol C8H18 as if it were 1 mol. Choice A (12.5 mol) inverts the ratio.

Spectator ions appear on both sides of the complete ionic equation unchanged. The complete ionic equation is: Cu2+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq) → Cu(OH)2(s) + 2Na+(aq) + SO42-(aq). Na+ and SO42- appear on both sides unchanged, making them spectator ions. Cu2+ and OH- actually participate in forming the precipitate Cu(OH)2, so they are not spectator ions.

First find moles of H2SO4: 0.100 mol/L x 0.0250 L = 0.00250 mol. Because the stoichiometric ratio is 1 mol H2SO4 to 2 mol NaOH, moles NaOH needed = 2 x 0.00250 = 0.00500 mol. Volume NaOH = 0.00500 mol / 0.250 mol/L = 0.0200 L = 20.0 mL. Choice C (25.0 mL) ignores the 1:2 stoichiometry and assumes a 1:1 ratio. Choice D (50.0 mL) applies the correct mole ratio but uses an incorrect calculation.

Moles PCl3 = 50.0 g / 137.33 g/mol = 0.3641 mol. The mole ratio is 1:1 (one mole PCl3 produces one mole PCl5), so moles PCl5 = 0.3641 mol. Mass PCl5 = 0.3641 mol x 208.24 g/mol = 75.8 g. Choice A (50.0 g) incorrectly assumes mass is conserved for this single product (it is not, since Cl2 mass is added). Choice C (104.1 g) doubles the result, corresponding to an incorrect 2:1 mole ratio.

For the nitrate ion, the sum of oxidation states equals the ion charge (-1). Each oxygen is -2, so: N + 3(-2) = -1, giving N - 6 = -1, thus N = +5. Choice A (+3) is the oxidation state of nitrogen in the nitrite ion (NO2-). Choice B (+4) is the oxidation state of nitrogen in nitrogen dioxide (NO2). Choice D (-3) is the oxidation state of nitrogen in ammonia (NH3) and ammonium ion (NH4+).

Mol H2 = 8.0 / 2.02 = 3.96 mol; mol O2 = 32.0 / 32.00 = 1.00 mol. The 2:1 ratio means 3.96 mol H2 requires 1.98 mol O2, but only 1.00 mol O2 is available. Therefore O2 is the limiting reagent. H2 consumed = 2 x 1.00 = 2.00 mol = 2.00 x 2.02 = 4.04 g. H2 remaining = 8.0 - 4.04 = 3.96 g, which rounds to 4.0 g. Choice A is wrong because there is significant excess H2. Choices C and D are wrong because O2 is completely consumed — it is the limiting reagent.

Using the half-reaction method: Oxidation half-reaction: Fe2+ → Fe3+ + e- (multiplied by 2). Reduction half-reaction: H2O2 + 2H+ + 2e- → 2H2O. Combining: 2Fe2+ + H2O2 + 2H+ → 2Fe3+ + 2H2O. The coefficient of Fe2+ is 2 and H2O2 is 1. Choice B is incorrect because one Fe2+ transfers only one electron, which cannot balance the two electrons gained by H2O2. Choice C doubles H2O2 unnecessarily, creating an electron imbalance.

From CO2: mol C = 1.466/44.01 = 0.03331 mol → 0.400 g C. From H2O: mol H2O = 0.600/18.02 = 0.03330 mol → 0.06660 mol H → 0.0671 g H. Mass O = 1.000 - 0.400 - 0.067 = 0.533 g → 0.0333 mol O. Mole ratio C:H:O = 0.0333:0.0666:0.0333 = 1:2:1. Empirical formula is CH2O (mass = 30.03 g/mol). Molecular mass / empirical mass = 60.05 / 30.03 = 2. Molecular formula = C2H4O2. Choice A (CH2O) is only the empirical formula with molar mass 30.03. Choice C (C2H6O, ethanol) has molar mass 46.07, not 60.05.

Mol NH3 = 34.0 / 17.03 = 1.997 mol, approximately 2.00 mol. In step 1, the NH3:NO mole ratio is 4:4 = 1:1, so 2.00 mol NH3 produces 2.00 mol NO. In step 2, the NO:NO2 ratio is 2:2 = 1:1, so 2.00 mol NO produces 2.00 mol NO2. Mass NO2 = 2.00 mol x 46.01 g/mol = 92.0 g. Choice A (46.0 g) corresponds to only 1.00 mol NO2, halving the result incorrectly. Choice D (184.0 g) doubles the answer, perhaps from misreading the molar ratio as 1:2.

Initial mol HCl = 0.05000 L x 1.000 M = 0.05000 mol. Mol NaOH used for excess HCl = 0.02000 L x 0.500 M = 0.01000 mol. Since HCl:NaOH = 1:1, excess mol HCl = 0.01000 mol. Mol HCl that reacted with MgO = 0.05000 - 0.01000 = 0.04000 mol. Since MgO:HCl = 1:2, mol MgO = 0.04000/2 = 0.02000 mol. Mass MgO = 0.02000 x 40.30 = 0.806 g. Percent purity = 0.806/1.500 x 100 = 53.7%. Choice D (80.6%) results from using the full 50.00 mL HCl without accounting for the back titration correction.

Balancing by half-reactions: Oxidation: SO3^2- + 2OH- → SO4^2- + H2O + 2e- (x3). Reduction: MnO4- + 2H2O + 3e- → MnO2 + 4OH- (x2). Combining and simplifying: 3SO3^2- + 6OH- + 2MnO4- + 4H2O → 3SO4^2- + 3H2O + 2MnO2 + 8OH-. Canceling common species: 3SO3^2- + 2MnO4- + H2O → 3SO4^2- + 2MnO2 + 2OH-. The 6 OH- from the oxidation half and 8 OH- from the reduction half simplify to a net 2 OH- on the products side. Choice A is wrong because OH- appears as a product, not a reactant, in the final balanced equation.

Mol CaCO3 = 25.0/100.09 = 0.2498 mol; mol HCl = 0.0400 L x 6.00 mol/L = 0.2400 mol. The 1:2 ratio means 0.2498 mol CaCO3 requires 0.4996 mol HCl, but only 0.2400 mol is available, so HCl is limiting. Mol CaCl2 from HCl = 0.2400/2 = 0.1200 mol. Theoretical yield = 0.1200 x 110.98 = 13.32 g. Actual yield = 0.850 x 13.32 = 11.3 g. Choice B (13.3 g) is the theoretical yield, not accounting for percent yield. Choices C and D assume CaCO3 is the limiting reagent, which is incorrect since HCl is limiting.

A synthesis reaction combines two or more substances to form a single product. Choice B shows sodium metal and chlorine gas combining to form one product, sodium chloride. Choice A is a decomposition reaction (one reactant breaks into multiple products). Choice C is a double displacement reaction (ions exchange partners). Choice D is a single displacement reaction (one element replaces another).

Most carbonates are insoluble, and calcium is not an exception — CaCO3 forms a precipitate when carbonate and calcium ions are combined in solution. NaNO3 is soluble because all nitrates are soluble and all sodium salts are soluble. (NH4)2SO4 is soluble because all ammonium salts are soluble. KOH is soluble because group 1 metal hydroxides are soluble.

Spectator ions appear unchanged on both sides of the complete ionic equation and do not participate in the net reaction. Na+ and NO3- remain as aqueous ions throughout. The driving force of the reaction is the formation of the AgCl precipitate, so Ag+ and Cl- are the active ions. The net ionic equation is simply Ag+(aq) + Cl-(aq) → AgCl(s), with Na+ and NO3- omitted as spectators.

To convert atoms to moles, divide by Avogadro's number: (9.03 × 10^23 atoms) / (6.022 × 10^23 atoms/mol) = 1.50 mol. One mole equals 6.022 × 10^23 particles, so 1.5 times that quantity gives 1.50 mol. A common error is multiplying instead of dividing by Avogadro's number, which would give an astronomically large number.

Starting with C3H8: the 3 carbons require 3 CO2, and the 8 hydrogens require 4 H2O. Counting oxygen atoms needed on the right: 3(2) + 4(1) = 10 oxygen atoms, which requires 5 O2 molecules. The balanced equation is C3H8 + 5O2 → 3CO2 + 4H2O. A coefficient of 4 is a common error when students only balance carbon and forget to account for hydrogen.

In H2SO4, hydrogen is +1 and oxygen is -2. Setting up the equation: 2(+1) + x + 4(-2) = 0 gives 2 + x - 8 = 0, so x = +6. Sulfur reaches its maximum oxidation state of +6 in sulfuric acid. The value +4 is a plausible distractor because sulfur commonly appears as +4 in SO2 (sulfur dioxide), but that compound has only two oxygen atoms, not four.

A reducing agent is the substance that is oxidized — it loses electrons. Zinc changes from oxidation state 0 to +2, losing two electrons, so Zn is oxidized and is therefore the reducing agent. H+ changes from +1 to 0 in H2, gaining electrons, making H+ the oxidizing agent. Cl- does not change oxidation state and acts as a spectator ion in this reaction. HCl as a compound is not the oxidizing agent — only the H+ ion within it participates in electron transfer.

Molar mass of H2O = 2(1.008) + 16.00 = 18.016 g/mol. Percent oxygen = (16.00 / 18.016) × 100 = 88.81%. Choice A (11.19%) is the percent by mass of hydrogen — a common mistake when the wrong element is used in the numerator. Choice C (66.67%) confuses the atom count fraction (1 out of 3 atoms) with the mass fraction, which is incorrect because atoms have different masses.

The net ionic equation includes only species that change during the reaction. K+ and NO3- remain as aqueous ions on both sides and are spectator ions that are cancelled out. Only Pb2+ and I- combine to form the PbI2 precipitate. Choice B is the complete molecular equation — it shows full formulas, not ions. Choice C is the complete ionic equation — all aqueous species are shown as ions, but spectators have not been cancelled. Choice D incorrectly shows a 1:1 product rather than the 1:2 stoichiometry required.

Percent yield = (actual yield / theoretical yield) × 100 = (6.97 g / 8.50 g) × 100 = 82.0%. The actual yield is always less than or equal to the theoretical yield in a real experiment due to side reactions, incomplete reactions, or losses during purification. Choice D (85.0%) results from a rounding or arithmetic error. Dividing incorrectly — placing theoretical in the numerator — would give a value greater than 100%, which is physically impossible.

A metal displaces another metal from solution only if it is higher in the activity series. Zinc is more active than silver, so zinc reduces Ag+ ions to silver metal while being oxidized to Zn2+. Choice A is incorrect because copper is less active than zinc and cannot displace it. Choice B is incorrect because silver is less active than hydrogen and will not displace H2 from acid. Choice D is incorrect because lead is less active than magnesium and cannot displace it.

mol N2 = 14.0 / 28.02 = 0.500 mol; mol H2 = 3.00 / 2.016 = 1.49 mol. To consume all N2, the reaction requires 0.500 × 3 = 1.50 mol H2. Only 1.49 mol H2 is available, so H2 runs out first — it is the limiting reagent. Choice C is a common error: having fewer moles does not automatically make a reagent limiting. The mole ratio from the balanced equation must always be used to compare amounts.

Weak acids are only partially ionized and must be written in molecular form in net ionic equations. Acetic acid remains as CH3COOH(aq), not split into H+ and CH3COO-. OH- from the strong base reacts with the intact weak acid molecule to produce water and the acetate ion. Choice A applies only to strong acid-strong base neutralizations, where the strong acid is fully ionized to H+. Choice D is the complete molecular equation, not the net ionic equation.

Assuming 100 g of compound: C = 40.0 g → 40.0 / 12.01 = 3.33 mol; H = 6.72 g → 6.72 / 1.008 = 6.67 mol; O = 53.3 g → 53.3 / 16.00 = 3.33 mol. Dividing all values by the smallest (3.33) gives C:H:O = 1:2:1, so the empirical formula is CH2O. Choice C (C2H4O2) has the same ratio but is not the simplest whole-number formula — it is exactly twice CH2O and represents the molecular formula of acetic acid or glycolaldehyde, not the empirical formula.

Using the dilution equation M1V1 = M2V2: (1.50 M)(V1) = (0.150 M)(500.0 mL), so V1 = (0.150 × 500.0) / 1.50 = 50.0 mL. The concentration decreases by a factor of 10 from stock to final solution, so the volume taken from stock must be 1/10 of the final volume: 500.0 / 10 = 50.0 mL. Choice D (150 mL) results from incorrectly using the ratio in the wrong direction.

mol Al = 5.40 / 26.98 = 0.200 mol. The balanced equation shows a 2:2 (1:1) mole ratio of Al to AlCl3, so mol AlCl3 = 0.200 mol. Mass AlCl3 = 0.200 mol × 133.34 g/mol = 26.7 g. Choice A (13.4 g) results from incorrectly using a 2:1 ratio instead of 1:1. Choice C (40.0 g) results from using the molar mass of Cl2 (70.90 g/mol) in the calculation instead of AlCl3.

In the reaction BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq), both reactants are ionic compounds that exchange their anions — Ba2+ pairs with SO4^2- and Na+ pairs with Cl-. This ion-swapping pattern is the defining feature of a double displacement (also called metathesis or exchange) reaction. Single displacement involves one element replacing another from a compound, which is not the case here. Synthesis requires simpler substances combining into one product.

Mn in MnO4- is +7; Mn2+ is +2 — a change of 5, requiring 5 electrons gained (reduction). Balance oxygen: add 4H2O to the right. Balance hydrogen: add 8H+ to the left. Check charge — left: (-1) + 8(+1) = +7; right: (+2) + 0 = +2. Adding 5e- to the left: +7 - 5 = +2. Balanced. Choice A incorrectly uses 3 electrons, which corresponds to a +7 → +4 change, not +7 → +2. Choice D shows electrons on the right, indicating oxidation — the wrong direction for a reduction half-reaction.

mol HCl initial = 0.200 M × 0.05000 L = 0.01000 mol. mol NaOH = 0.100 M × 0.01000 L = 1.00 × 10^-3 mol. Since HCl and NaOH react 1:1, mol excess HCl = 1.00 × 10^-3 mol. mol HCl that reacted with CaCO3 = 0.01000 - 0.001000 = 0.009000 mol. Since 2 mol HCl reacts per mol CaCO3: mol CaCO3 = 0.009000 / 2 = 4.50 × 10^-3 mol. Mass CaCO3 = 4.50 × 10^-3 × 100.09 = 0.4504 g. Percent purity = (0.4504 / 0.500) × 100 = 90.1%. A common error is forgetting the 2:1 HCl:CaCO3 stoichiometry, which doubles the calculated CaCO3 and gives an impossible purity over 100%.

mol SO2 = 64.1 / 64.07 = 1.00 mol. Step 1 (1:1 ratio): theoretical SO3 = 1.00 mol; actual SO3 = 1.00 × 0.850 = 0.850 mol. Step 2 (1:1 ratio): theoretical H2SO4 from 0.850 mol SO3 = 0.850 mol; actual H2SO4 = 0.850 × 0.900 = 0.765 mol. Mass = 0.765 × 98.08 = 75.0 g. Choice A (37.5 g) results from misreading the 2:2 ratio in Step 1 as 2:1, halving the SO3 produced. Choice C (83.4 g) applies only the 85.0% yield from Step 1, ignoring Step 2. Choice D (98.1 g) is the 100% theoretical yield with no corrections applied.

Oxidation half-reaction: Cr goes +3 → +6 (loses 3e-). Balance O by adding H2O, then H using H+, then convert to basic with OH-: Cr(OH)3 + 5OH- → CrO4^2- + 4H2O + 3e-. Reduction half-reaction: Cl2 + 2e- → 2Cl-. To balance electrons, multiply oxidation by 2 and reduction by 3 (LCM = 6): 2Cr(OH)3 + 10OH- → 2CrO4^2- + 8H2O + 6e- and 3Cl2 + 6e- → 6Cl-. Overall: 2Cr(OH)3 + 10OH- + 3Cl2 → 2CrO4^2- + 8H2O + 6Cl-. The coefficient of OH- is 10. Choice C (8) is the coefficient of H2O, a common confusion.

mol C = mol CO2 = 1.571 / 44.01 = 0.03570 mol. mol H = 2 × mol H2O = 2 × (0.643 / 18.02) = 0.07136 mol. Ratio C:H = 0.03570 : 0.07136 = 1 : 2.00, giving empirical formula CH2 (molar mass = 14.03 g/mol). Dividing the given molar mass: 58 / 14.03 ≈ 4. The molecular formula is 4 × CH2 = C4H8. Choice C (C3H6, molar mass = 42 g/mol) has the correct empirical ratio but does not match the given molar mass of 58 g/mol. Choice A is only the empirical formula, not the molecular formula.

mol NH3 = 34.0 / 17.03 = 2.00 mol; mol O2 = 80.0 / 32.00 = 2.50 mol. Check limiting reagent: 2.00 mol NH3 requires 2.00 × (5/4) = 2.50 mol O2 — exactly the amount present, so both reactants are consumed simultaneously. Theoretical NO = 2.00 × (4/4) = 2.00 mol. With 80.0% yield: actual NO = 2.00 × 0.800 = 1.60 mol. Mass = 1.60 × 30.01 = 48.0 g. Choice D (60.0 g) is the theoretical yield without applying the percent yield. Choice B (38.4 g) results from applying the 80% yield factor twice.

mol NaOH = 0.150 mol/L × 0.0333 L = 5.00 × 10^-3 mol. The reaction requires 2 mol NaOH per mol H2A, so mol H2A = 5.00 × 10^-3 / 2 = 2.50 × 10^-3 mol. Molar mass = 0.250 g / 2.50 × 10^-3 mol = 100 g/mol. Choice A (50.1 g/mol) results from a very common error: forgetting the 2:1 stoichiometry and using mol NaOH directly as mol H2A. Failing to account for the diprotic nature of the acid halves the calculated molar mass.

mol Cr2O7^2- = 0.01855 L × 0.0200 mol/L = 3.71 × 10^-4 mol. The balanced equation shows a 1:6 mole ratio of Cr2O7^2- to Fe2+, so mol Fe2+ = 6 × 3.71 × 10^-4 = 2.226 × 10^-3 mol. Molarity of Fe2+ = 2.226 × 10^-3 mol / 0.02500 L = 0.0890 M. Choice B (0.0445 M) results from using a 1:3 ratio instead of 1:6. Choice A (0.0148 M) results from using a 1:1 ratio, ignoring stoichiometry entirely. Careful attention to the mole ratio in redox titrations is essential because it is rarely 1:1.