Chemistry Practice Test 2: Stoichiometry
- Stoichiometry
- 10 questions
- 15 minutes
This second stoichiometry test uses balanced equations to predict amounts of reactants and products. Use these rounded atomic masses: H = 1, C = 12, N = 14, O = 16, Mg = 24, Ca = 40, Zn = 65. Tip: convert grams to moles first, use the mole ratio, then convert back.
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Question 1
Potassium chlorate decomposes: 2KClO3 → 2KCl + 3O2. How many moles of O2 form from 4.0 mol of KClO3?
Correct answer: B
The mole ratio is 3 mol O2 per 2 mol KClO3, so 4.0 × 3/2 = 6.0 mol. Using the ratio upside down gives 4.0 × 2/3 ≈ 2.7 mol.
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Question 2
Methane burns: CH4 + 2O2 → CO2 + 2H2O. What mass of CO2 forms when 16 g of CH4 burns completely?
Correct answer: C
16 g of CH4 is 1 mol, which gives 1 mol of CO2, or 44 g. The answer 16 g assumes mass is conserved for each substance separately, which it is not.
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Question 3
Methane burns: CH4 + 2O2 → CO2 + 2H2O. What mass of O2 is needed to burn 16 g of CH4 completely?
Correct answer: A
1 mol of CH4 needs 2 mol of O2, and 2 × 32 g = 64 g. The answer 32 g forgets the coefficient 2 in front of O2.
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Question 4
Hydrogen and oxygen react: 2H2 + O2 → 2H2O. A mixture contains 3.0 mol of H2 and 2.0 mol of O2. Which reactant is limiting, and how much water forms?
Correct answer: D
3.0 mol of H2 needs only 1.5 mol of O2, and 2.0 mol is available, so H2 runs out first and makes 3.0 mol of water. O2 would need 4.0 mol of H2, which is more than there is.
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Question 5
Ammonia is made by N2 + 3H2 → 2NH3. A reaction starts with 28 g of N2 and 9.0 g of H2. What is the maximum mass of NH3 that can form?
Correct answer: A
28 g of N2 is 1.0 mol and 9.0 g of H2 is 4.5 mol; 1.0 mol of N2 needs only 3.0 mol of H2, so N2 is limiting and gives 2.0 mol (34 g) of NH3. 51 g is what the H2 could make if there were enough N2.
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Question 6
In the reaction N2 + 3H2 → 2NH3, 28 g of N2 reacts completely with some of the 9.0 g of H2 present. What mass of H2 is left over?
Correct answer: B
1.0 mol of N2 uses 3.0 mol (6.0 g) of H2, so 9.0 − 6.0 = 3.0 g is left. 6.0 g is the amount used, not the amount left, and 1.5 g is the leftover in moles.
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Question 7
Heating 10.0 g of CaCO3 should produce CO2 by CaCO3 → CaO + CO2. A student collects 3.3 g of CO2. What is the percent yield?
Correct answer: D
10.0 g of CaCO3 is 0.100 mol, so the theoretical yield is 0.100 × 44 = 4.4 g of CO2, and 3.3 ÷ 4.4 × 100 = 75%. 33% compares 3.3 g with the 10.0 g of reactant instead of the theoretical yield.
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Question 8
Magnesium burns: 2Mg + O2 → 2MgO. A student burns 4.8 g of Mg, and the reaction has a 90% yield. What mass of MgO does the student collect?
Correct answer: C
4.8 g of Mg is 0.20 mol, which gives 0.20 mol of MgO, or 0.20 × 40 = 8.0 g in theory; 90% of 8.0 g is 7.2 g. 4.3 g takes 90% of the magnesium mass instead of the product mass.
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Question 9
Aluminum reacts with chlorine: 2Al + 3Cl2 → 2AlCl3. How many moles of Cl2 are needed to react with 0.60 mol of Al?
Correct answer: D
The ratio is 3 mol Cl2 per 2 mol Al, so 0.60 × 3/2 = 0.90 mol. Using 2/3 instead gives 0.40 mol.
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Question 10
Zinc reacts with hydrochloric acid: Zn + 2HCl → ZnCl2 + H2. At STP, 1 mol of gas occupies 22.4 L. What volume of H2 at STP forms from 13 g of zinc with excess acid?
Correct answer: B
13 g ÷ 65 g/mol = 0.20 mol of Zn, which gives 0.20 mol of H2, and 0.20 × 22.4 L = 4.48 L. 8.96 L wrongly uses the coefficient 2 from HCl.
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