Theoretical Yield Practice Problems

The most common error in stoichiometry is treating the theoretical yield as something you can measure, which is why theoretical yield practice problems are so important. It is not. The theoretical yield is the mass of product you calculate from the limiting reactant, assuming perfect conversion, no side reactions, and zero loss. It is a ceiling, not a target. Real reactions almost never hit it. These problems close that gap between calculation and lab reality, covering every type you will see in high school and first-year college chemistry. Each problem works from a balanced equation, uses CIAAW standard atomic weights, and is verified against the site calculator. Worked solutions are collapsible so you try the problem before you check the answer.

When one reactant mass is given and the others are in excess, the path is mass → moles → mole ratio → moles of product → mass of product. The failure mode is skipping the mole conversion. Divide grams of reactant by molar mass to get moles; multiply by the mole ratio from the balanced equation; multiply by the molar mass of the product.

Problem 1: Iron(III) oxide reacts with carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. If 25.0 g of iron(III) oxide is used, what is the theoretical yield of iron? (Molar masses: Fe₂O₃ = 159.69 g/mol, Fe = 55.85 g/mol)

Problem 2: Ammonia is produced from nitrogen and hydrogen: N₂ + 3H₂ → 2NH₃. If 10.0 g of hydrogen gas is reacted with excess nitrogen, what is the theoretical yield of ammonia? (Molar masses: H₂ = 2.016 g/mol, NH₃ = 17.031 g/mol)

Problem 3: Sodium chloride is produced from sodium and chlorine: 2Na + Cl₂ → 2NaCl. If 5.00 g of sodium is used, what is the theoretical yield of sodium chloride? (Molar masses: Na = 22.99 g/mol, NaCl = 58.44 g/mol)

Find the Reagent That Runs Out First

The limiting reagent is the one that produces the least product. Convert both starting masses to moles, then to moles of the same product using the mole ratio. The reagent that yields the smaller product amount is the limit. Excess reactant remains after the reaction.

Problem 4: Aluminum reacts with oxygen: 4Al + 3O₂ → 2Al₂O₃. If 10.0 g of aluminum and 10.0 g of oxygen are reacted, which is the limiting reactant and what is the theoretical yield of aluminum oxide? (Molar masses: Al = 26.98 g/mol, O₂ = 32.00 g/mol, Al₂O₃ = 101.96 g/mol)

Problem 5: Methane burns in oxygen: CH₄ + 2O₂ → CO₂ + 2H₂O. If 8.00 g of methane and 20.0 g of oxygen are mixed, what is the limiting reactant and the theoretical yield of carbon dioxide? (Molar masses: CH₄ = 16.04 g/mol, O₂ = 32.00 g/mol, CO₂ = 44.01 g/mol)

Problem 6: Zinc reacts with hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. If 5.00 g of zinc and 10.0 g of hydrochloric acid are reacted, which reactant is limiting and what mass of zinc chloride is produced? (Molar masses: Zn = 65.38 g/mol, HCl = 36.46 g/mol, ZnCl₂ = 136.29 g/mol)

Solution stoichiometry uses molarity (mol/L) and volume in liters. Moles = molarity × volume (L). The rest follows the same mass → moles → product path. The common error is using volume in mL without converting to L.

Problem 7: Lead(II) nitrate and potassium iodide react: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃. If 25.0 mL of 0.200 M lead(II) nitrate is mixed with 30.0 mL of 0.300 M potassium iodide, what is the theoretical yield of lead(II) iodide? (Molar mass: PbI₂ = 461.0 g/mol)

Problem 8: Silver nitrate reacts with sodium chloride: AgNO₃ + NaCl → AgCl + NaNO₃. If 15.0 mL of 0.100 M silver nitrate is added to 20.0 mL of 0.150 M sodium chloride, what mass of silver chloride precipitate forms? (Molar mass: AgCl = 143.32 g/mol)

Gas stoichiometry uses the IUPAC definition of STP: 273.15 K and 100 kPa. At STP, one mole of an ideal gas occupies 22.710 L (the molar volume). For gases not at STP, use PV = nRT. The convention used here is the IUPAC definition, not the older 0 °C and 1 atm (101.325 kPa, 22.4 L/mol).

Problem 9: Magnesium reacts with hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. What volume of hydrogen gas at STP is produced from 2.43 g of magnesium? (Molar mass: Mg = 24.31 g/mol)

Problem 10: Calcium carbonate decomposes: CaCO₃ → CaO + CO₂. What volume of carbon dioxide at STP is produced from 10.0 g of calcium carbonate? (Molar mass: CaCO₃ = 100.09 g/mol)

Reagents are rarely 100% pure. The mass of pure compound = mass of sample × (purity / 100). Use the pure mass for stoichiometric calculations. A bottle labeled 95% purity means only 95% of the weighed mass is the actual reactant.

Problem 11: A 5.00 g sample of impure calcium carbonate (CaCO₃, 80% purity) reacts with excess hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. What is the theoretical yield of carbon dioxide? (Molar masses: CaCO₃ = 100.09 g/mol, CO₂ = 44.01 g/mol)

Problem 12: A 10.0 g sample of impure zinc (Zn, 90% purity) is reacted with excess sulfuric acid: Zn + H₂SO₄ → ZnSO₄ + H₂. What volume of hydrogen gas at STP is produced? (Molar mass: Zn = 65.38 g/mol)

Compare What You Got to What You Could Have Gotten

Percent yield = (actual yield / theoretical yield) × 100%. The theoretical yield comes from the limiting reactant calculation; the actual yield is the mass recovered on the balance. A 95% yield is excellent in most organic syntheses; a 50% yield can be acceptable for a difficult multistep route. If your actual yield exceeds theoretical, check for wet or impure product.

Problem 13: In a reaction that produces water from hydrogen and oxygen, the theoretical yield of water is 18.0 g. If 15.3 g of water is recovered, what is the percent yield?

Problem 14: The theoretical yield of aspirin from a synthesis is 5.00 g. The actual yield is 3.85 g. What is the percent yield?

Problem 15: A student performs a reaction with a theoretical yield of 12.5 g of copper. They recover 10.2 g. What is the percent yield, and is this considered good or moderate?

Percent Yield Range	Common Lab Description	Typical Context
< 40%	Low	Unoptimized reaction, significant losses, or measurement error
40–60%	Moderate	Acceptable for first attempts or difficult syntheses
60–80%	Good	Solid result for most lab procedures
80–90%	Excellent	Well-optimized reaction, minimal losses
> 90%	Near Quantitative	Exceptional; check for calculation errors if over 100%

Problem 1 Answer

Mole ratio Fe₂O₃:Fe = 1:2. Moles Fe = 0.1565 × 2 = 0.3130 mol. Mass Fe = 0.3130 mol × 55.85 g/mol = 17.5 g.

Problem 2 Answer

Moles H₂ = 10.0 g / 2.016 g/mol = 4.960 mol. Mole ratio H₂:NH₃ = 3:2. Moles NH₃ = 4.960 × (2/3) = 3.307 mol. Mass NH₃ = 3.307 mol × 17.031 g/mol = 56.3 g.

Problem 3 Answer

Moles Na = 5.00 g / 22.99 g/mol = 0.2175 mol. Mole ratio Na:NaCl = 2:2 = 1:1. Moles NaCl = 0.2175 mol. Mass NaCl = 0.2175 mol × 58.44 g/mol = 12.7 g.

Problem 4 Answer

Moles O₂ = 10.0 g / 32.00 g/mol = 0.3125 mol.O₂ to Al₂O₃: 0.3125 mol O₂ × (2/3) = 0.2083 mol Al₂O₃. Al is limiting. Theoretical yield Al₂O₃ = 0.1854 mol × 101.96 g/mol = 18.9 g.

Problem 5 Answer

Moles CH₄ = 8.00 g / 16.04 g/mol = 0.4988 mol. Moles O₂ = 20.0 g / 32.00 g/mol = 0.6250 mol. CH₄ to CO₂: 0.4988 mol × (1/1) = 0.4988 mol CO₂. O₂ to CO₂: 0.6250 mol × (1/2) = 0.3125 mol CO₂. O₂ is limiting. Theoretical yield CO₂ = 0.3125 mol × 44.01 g/mol = 13.8 g.

Problem 6 Answer

Moles Zn = 5.00 g / 65.38 g/mol = 0.07648 mol. Moles HCl = 10.0 g / 36.46 g/mol = 0.2743 mol. Zn to ZnCl₂: 0.07648 mol Zn × (1/1) = 0.07648 mol ZnCl₂. HCl to ZnCl₂: 0.2743 mol HCl × (1/2) = 0.1372 mol ZnCl₂. Zn is limiting. Theoretical yield ZnCl₂ = 0.07648 mol × 136.29 g/mol = 10.4 g.

Problem 7 Answer

Moles Pb(NO₃)₂ = 0.200 M × 0.0250 L = 0.00500 mol. Moles KI = 0.300 M × 0.0300 L = 0.00900 mol. Pb(NO₃)₂ to PbI₂: 0.00500 mol × (1/1) = 0.00500 mol PbI₂. KI to PbI₂: 0.00900 mol × (1/2) = 0.00450 mol PbI₂. KI is limiting. Theoretical yield PbI₂ = 0.00450 mol × 461.0 g/mol = 2.07 g.

Problem 8 Answer

Moles AgNO₃ = 0.100 M × 0.0150 L = 0.00150 mol. Moles NaCl = 0.150 M × 0.0200 L = 0.00300 mol. Mole ratio 1:1. AgNO₃ is limiting (0.00150 mol < 0.00300 mol). Moles AgCl = 0.00150 mol. Mass AgCl = 0.00150 mol × 143.32 g/mol = 0.215 g.

Problem 9 Answer

Moles Mg = 2.43 g / 24.31 g/mol = 0.1000 mol. Mole ratio Mg:H₂ = 1:1. Moles H₂ = 0.1000 mol. Volume at STP = 0.1000 mol × 22.710 L/mol = 2.27 L.

Problem 10 Answer

Moles CaCO₃ = 10.0 g / 100.09 g/mol = 0.09991 mol. Mole ratio CaCO₃:CO₂ = 1:1. Moles CO₂ = 0.09991 mol. Volume at STP = 0.09991 mol × 22.710 L/mol = 2.27 L.

Problem 11 Answer

Pure CaCO₃ = 5.00 g × 0.80 = 4.00 g. Moles CaCO₃ = 4.00 g / 100.09 g/mol = 0.03996 mol. Mole ratio CaCO₃:CO₂ = 1:1. Moles CO₂ = 0.03996 mol. Mass CO₂ = 0.03996 mol × 44.01 g/mol = 1.76 g.

Problem 12 Answer

Pure Zn = 10.0 g × 0.90 = 9.00 g. Moles Zn = 9.00 g / 65.38 g/mol = 0.1377 mol. Mole ratio Zn:H₂ = 1:1. Moles H₂ = 0.1377 mol. Volume at STP = 0.1377 mol × 22.710 L/mol = 3.13 L.

Problem 13 Answer

Percent yield = (15.3 g / 18.0 g) × 100% = 85.0%.

Problem 14 Answer

Percent yield = (3.85 g / 5.00 g) × 100% = 77.0%.

Problem 15 Answer

Percent yield = (10.2 g / 12.5 g) × 100% = 81.6%. According to the table, this is considered excellent.