Rucete ✏ AP Chemistry In a Nutshell
5. Stoichiometry — Practice Questions
This chapter introduces stoichiometric methods, including dimensional analysis, percent composition, empirical formulas, limiting-reactant calculations, and titration-based molarity analysis.
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(Multiple Choice — Click to Reveal Answer)
1. How many moles of water are produced when 2.0 mol of hydrogen gas reacts with 1.0 mol of oxygen gas?
(A) 1.0 mol
(B) 2.0 mol
(C) 3.0 mol
(D) 4.0 mol
Answer
(B) — According to 2H₂ + O₂ → 2H₂O, 2 mol H₂ produces 2 mol H₂O.
2. A 0.500 L solution contains 0.250 mol of KNO₃. What is the molarity of the solution?
(A) 0.125 M
(B) 0.250 M
(C) 0.500 M
(D) 1.00 M
Answer
(C) — M = mol/L = 0.250 mol / 0.500 L = 0.500 M.
3. How many molecules are there in 0.25 mol of CO₂?
(A) 1.51 × 10²³
(B) 6.02 × 10²³
(C) 3.01 × 10²³
(D) 1.00 × 10²⁴
Answer
(C) — 0.25 mol × (6.02 × 10²³) = 1.505 × 10²³ ≈ 3.01 × 10²³.
4. What is the percent composition of nitrogen in NH₄NO₃?
(A) 14.0%
(B) 17.5%
(C) 35.0%
(D) 58.3%
Answer
(C) — NH₄NO₃ has two nitrogen atoms: (28.0 g / 80.0 g) × 100 = 35.0%.
5. Which of the following compounds has the empirical formula CH₂?
(A) C₂H₄
(B) C₄H₈
(C) C₆H₁₂
(D) All of the above
Answer
(D) — All reduce to the same empirical ratio CH₂.
6. In a titration, 25.0 mL of 0.150 M NaOH is required to neutralize 50.0 mL of HCl. What is the concentration of HCl?
(A) 0.0750 M
(B) 0.150 M
(C) 0.300 M
(D) 0.600 M
Answer
(A) — M₁V₁ = M₂V₂ → 0.150 × 25.0 = M × 50.0 → M = 0.0750 M.
7. Which quantity is conserved in all chemical reactions?
(A) Mass only
(B) Moles only
(C) Volume only
(D) Both mass and charge
Answer
(D) — The law of conservation of mass and charge always applies.
8. What is the limiting reactant if 10.0 g of H₂ reacts with 80.0 g of O₂ to form water?
(A) H₂
(B) O₂
(C) H₂O
(D) Not enough information
Answer
(B) — Convert both to moles: 10 g H₂ = 5 mol; 80 g O₂ = 2.5 mol → H₂ is in excess.
9. What mass of NaCl is formed when 0.100 mol of HCl reacts with NaOH?
(A) 5.85 g
(B) 11.7 g
(C) 2.93 g
(D) 0.585 g
Answer
(A) — 0.100 mol NaCl × 58.5 g/mol = 5.85 g.
10. A gas occupies 44.8 L at STP. How many moles of the gas are present?
(A) 1 mol
(B) 2 mol
(C) 0.5 mol
(D) 4 mol
Answer
(B) — 44.8 L / 22.4 L/mol = 2 mol.
11. How many grams are in 0.250 mol of CaCO₃ (Molar mass = 100.1 g/mol)?
(A) 25.0 g
(B) 50.0 g
(C) 75.1 g
(D) 100.1 g
Answer
(B) — 0.250 mol × 100.1 g/mol = 25.025 g ≈ 25.0 g.
12. What is the molar mass of C₆H₁₂O₆?
(A) 90.0 g/mol
(B) 120.0 g/mol
(C) 180.0 g/mol
(D) 210.0 g/mol
Answer
(C) — (6 × 12.0) + (12 × 1.0) + (6 × 16.0) = 180.0 g/mol.
13. What is the empirical formula of a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass?
(A) CHO
(B) CH₂O
(C) C₂H₄O₂
(D) C₃H₆O₃
Answer
(B) — The simplest ratio for these percentages corresponds to CH₂O.
14. Which of the following is a correct balanced equation for the combustion of methane (CH₄)?
(A) CH₄ + O₂ → CO₂ + H₂O
(B) CH₄ + 2O₂ → CO₂ + 2H₂O
(C) CH₄ + 3O₂ → CO + 2H₂O
(D) CH₄ + O₂ → C + 2H₂
Answer
(B) — CH₄ + 2O₂ → CO₂ + 2H₂O is the balanced combustion reaction.
15. How many oxygen atoms are in 2.0 mol of SO₄²⁻ ions?
(A) 2.0 × 10²³
(B) 4.0 × 10²³
(C) 4.8 × 10²⁴
(D) 8.0 × 10²³
Answer
(C) — 2 mol × 4 atoms/mol × 6.02 × 10²³ = 4.82 × 10²⁴ atoms.
16. What is the limiting reactant in a reaction of 4.00 g of H₂ with 32.0 g of O₂ to produce water?
(A) H₂
(B) O₂
(C) H₂O
(D) Cannot be determined
Answer
(B) — 4 g H₂ = 2 mol; 32 g O₂ = 1 mol. Needs 2 mol H₂ per mol O₂ → O₂ limits.
17. If 5.00 g of Ca reacts with excess HCl, how many liters of H₂ gas are produced at STP?
Ca + 2HCl → CaCl₂ + H₂
(A) 1.12 L
(B) 2.24 L
(C) 3.36 L
(D) 4.48 L
Answer
(B) — 5.00 g Ca = 0.125 mol → 0.125 mol H₂ × 22.4 L/mol = 2.80 L ≈ 2.24 L.
18. What mass of H₂O is produced from the combustion of 1 mol of propane (C₃H₈)?
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
(A) 18.0 g
(B) 36.0 g
(C) 72.0 g
(D) 90.0 g
Answer
(C) — 4 mol H₂O × 18.0 g/mol = 72.0 g.
19. Which of the following represents a correct mole ratio from the equation: 2Al + 3Cl₂ → 2AlCl₃?
(A) 2 mol Al / 3 mol Cl₂
(B) 3 mol Cl₂ / 2 mol AlCl₃
(C) 1 mol Al / 1 mol AlCl₃
(D) All of the above
Answer
(D) — All are valid mole ratios from the balanced equation.
20. How many grams of CO₂ are produced from 100.0 g of C₂H₆ combustion?
C₂H₆ + (7/2)O₂ → 2CO₂ + 3H₂O
(Molar masses: C₂H₆ = 30.0 g/mol, CO₂ = 44.0 g/mol)
(A) 88.0 g
(B) 146.7 g
(C) 293.3 g
(D) 220.0 g
Answer
(B) — 100 g C₂H₆ ÷ 30 = 3.33 mol → 2 mol CO₂ per mol C₂H₆ → 6.67 mol CO₂ × 44 = 293.3 g.
21. What is the mole ratio of H₂O to O₂ in the reaction: 2H₂ + O₂ → 2H₂O?
(A) 1:1
(B) 2:1
(C) 1:2
(D) 2:2
Answer
(B) — 2 mol H₂O are produced for every 1 mol O₂ consumed.
22. A sample of gas has a volume of 11.2 L at STP. How many moles of gas are present?
(A) 0.25 mol
(B) 0.50 mol
(C) 1.00 mol
(D) 2.00 mol
Answer
(B) — 11.2 L ÷ 22.4 L/mol = 0.50 mol.
23. What is the volume (at STP) of 3.00 mol of oxygen gas?
(A) 22.4 L
(B) 44.8 L
(C) 67.2 L
(D) 89.6 L
Answer
(C) — 3 mol × 22.4 L/mol = 67.2 L.
24. Which term refers to the ratio of actual yield to theoretical yield, expressed as a percentage?
(A) Limiting yield
(B) Percent yield
(C) Theoretical efficiency
(D) Stoichiometric ratio
Answer
(B) — Percent yield = (actual / theoretical) × 100.
25. Which compound contains the greatest percentage by mass of oxygen?
(A) CO₂
(B) H₂O
(C) CH₃OH
(D) C₂H₅OH
Answer
(A) — CO₂ has two oxygen atoms and a relatively small carbon atom, leading to ~72% oxygen by mass.
26. In a reaction where 2.00 mol of A reacts with 3.00 mol of B to form product C, the balanced equation is:
2A + 2B → C.
What is the limiting reactant?
(A) A
(B) B
(C) C
(D) Cannot be determined
Answer
(B) — 2 mol A needs 2 mol B. Since 3 mol B is more than required, B is in excess → A limits.
27. How many moles of oxygen are required to completely combust 2.50 mol of ethanol (C₂H₅OH)?
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
(A) 2.50 mol
(B) 5.00 mol
(C) 6.25 mol
(D) 7.50 mol
Answer
(D) — 2.50 mol ethanol × 3 mol O₂ / 1 mol ethanol = 7.50 mol O₂.
28. A compound contains 70% iron and 30% oxygen by mass. What is the empirical formula?
(A) Fe₂O₃
(B) FeO
(C) Fe₃O₄
(D) Fe₂O
Answer
(A) — 70 g Fe / 55.8 = 1.25 mol, 30 g O / 16 = 1.875 mol → ratio ~2:3 → Fe₂O₃.
29. Which reagent is in excess when 10.0 g of Al reacts with 50.0 g of Cl₂?
2Al + 3Cl₂ → 2AlCl₃
(A) Al
(B) Cl₂
(C) AlCl₃
(D) Neither
Answer
(B) — 10 g Al = 0.37 mol, 50 g Cl₂ = 0.70 mol → needs 0.55 mol Cl₂ → excess.
30. What is the theoretical yield (in grams) of CO₂ from the combustion of 44.0 g of propane (C₃H₈)?
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
(Molar masses: C₃H₈ = 44.0 g/mol, CO₂ = 44.0 g/mol)
(A) 44.0 g
(B) 88.0 g
(C) 132.0 g
(D) 176.0 g
Answer
(C) — 1 mol C₃H₈ gives 3 mol CO₂ → 3 × 44 = 132.0 g CO₂.
31. How many grams of water will be formed from 40.0 g of H₂ and excess O₂?
2H₂ + O₂ → 2H₂O
(A) 72.0 g
(B) 180.0 g
(C) 360.0 g
(D) 80.0 g
Answer
(A) — 40 g H₂ = 20 mol → 20 mol H₂O × 18 g/mol = 360 g.
32. In a lab, 2.50 g of product is obtained from a reaction with a theoretical yield of 3.00 g. What is the percent yield?
(A) 83.3%
(B) 88.5%
(C) 91.2%
(D) 93.8%
Answer
(A) — (2.50 / 3.00) × 100 = 83.3%.
33. A sample contains 4.0 g of hydrogen and 32.0 g of oxygen. What is the empirical formula?
(A) H₂O
(B) H₂O₂
(C) HO
(D) H₄O₄
Answer
(A) — 4.0 g H = 4 mol, 32.0 g O = 2 mol → ratio 2:1 → H₂O.
34. Which volume of 0.250 M H₂SO₄ is needed to neutralize 100.0 mL of 0.500 M NaOH?
(A) 25.0 mL
(B) 50.0 mL
(C) 100.0 mL
(D) 200.0 mL
Answer
(B) — H₂SO₄:NaOH = 1:2, so M₁V₁ = M₂V₂ / 2 → (0.250)V = 0.500 × 100 / 2 → V = 50.0 mL.
35. When 5.00 g of Na reacts with excess water, how many grams of H₂ are produced?
2Na + 2H₂O → 2NaOH + H₂
(Molar masses: Na = 23.0, H₂ = 2.0)
(A) 0.109 g
(B) 0.217 g
(C) 0.435 g
(D) 1.09 g
Answer
(C) — 5.00 g Na / 23 = 0.217 mol → 0.109 mol H₂ × 2.0 g = 0.435 g.
36. Explain how to determine the limiting reactant when given masses of two reactants.
Answer
Convert both reactant masses to moles, then use the mole ratio from the balanced equation to find which produces fewer moles of product — that one is the limiting reactant.
37. A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen. Determine the empirical formula.
Answer
C₂H₆O — Convert percentages to moles: C: 52.2/12.0 ≈ 4.35, H: 13.0/1.0 = 13.0, O: 34.8/16 ≈ 2.18 → divide all by 2.18 → ≈ C₂H₆O.
38. Describe the steps to calculate percent composition of a compound.
Answer
Step 1: Find the molar mass of the compound. Step 2: Divide the mass of each element by total molar mass. Step 3: Multiply by 100 to get percent.
39. What is the theoretical yield of Al₂O₃ when 10.0 g of Al reacts with excess O₂? 4Al + 3O₂ → 2Al₂O₃
Answer
18.9 g — 10.0 g Al = 0.370 mol → 0.185 mol Al₂O₃ × 102 g/mol = 18.9 g.
40. Explain why percent yield is always less than 100% in a real reaction.
Answer
Due to losses during transfer, side reactions, incomplete reactions, or measurement errors.
41. How can you determine the molar mass of a gas at STP using its volume?
Answer
Use the volume (L) and convert to moles using 22.4 L/mol, then divide given mass by moles to get molar mass.
42. What mass of glucose (C₆H₁₂O₆) contains 0.500 mol of the compound?
Answer
90.0 g — 0.500 mol × 180 g/mol = 90.0 g.
43. A 1.00 L solution contains 0.500 mol of NaCl. What is the molarity?
Answer
0.500 M — M = mol / L = 0.500 mol / 1.00 L.
44. Describe the role of the mole ratio in stoichiometric calculations.
Answer
It relates the amount of reactants to products using coefficients from the balanced chemical equation.
45. What is the volume of 0.750 mol of a gas at STP?
Answer
16.8 L — 0.750 mol × 22.4 L/mol = 16.8 L.
46. Why must chemical equations be balanced before doing stoichiometry?
Answer
To satisfy the law of conservation of mass, and to ensure correct mole ratios between substances.
47. A 25.0 mL sample of HCl requires 30.0 mL of 0.100 M NaOH to reach the endpoint. What is the concentration of HCl?
Answer
0.120 M — M₁V₁ = M₂V₂ → M × 25.0 = 0.100 × 30.0 → M = 0.120 M.
48. What is the percent composition of oxygen in H₂O?
Answer
88.9% — (16.0 / 18.0) × 100 = 88.9%
49. A sample contains 0.10 mol of a compound. If its mass is 7.80 g, what is its molar mass?
Answer
78.0 g/mol — Molar mass = mass / mol = 7.80 g / 0.10 mol.
50. How many water molecules are in 18.0 g of H₂O?
Answer
6.02 × 10²³ — 18.0 g = 1 mol → 1 mol = Avogadro's number of molecules.