Stoichiometry & Yield — Chapter review · Physical Science: Chemistry, Grade 11

Calculate reacting amounts, identify shortages, and interpret production results.

Review the key ideas

Stoichiometry

The coefficients of a balanced equation give mole ratios (not mass ratios); a mass-to-mass problem follows one road map — mass A → moles A → (× mole ratio) → moles B → mass B.

Limiting Reactants & Percent Yield

The limiting reactant runs out first and sets the theoretical yield (the maximum product possible); percent yield = (actual yield ÷ theoretical yield) × 100% measures how much of that maximum a real reaction actually delivers.

Solution Stoichiometry

Convert solution volume to moles before using the reaction ratio.

Problem 1

For 2H₂ + O₂ → 2H₂O, how many moles of oxygen (O₂) are needed to react completely with 4.0 mol of hydrogen (H₂)?

Answer and explanation

2.0 mol O₂

The mole ratio of H₂ to O₂ is 2 : 1. So 4.0 mol H₂ × (1 mol O₂ ÷ 2 mol H₂) = 2.0 mol O₂.

Problem 2

What mass of water is produced when 8.08 g of hydrogen reacts with excess oxygen? 2H₂ + O₂ → 2H₂O. (H₂ = 2.02 g/mol, H₂O = 18.02 g/mol.)

Answer and explanation

72.08 g

Convert to moles: 8.08 g ÷ 2.02 g/mol = 4.00 mol H₂. Apply the mole ratio H₂O : H₂ = 2 : 2 = 1 : 1, giving 4.00 mol H₂O. Convert back to mass: 4.00 mol × 18.02 g/mol = 72.08 g.

Problem 3

A reaction has a theoretical yield of 25.0 g but produces an actual yield of 22.0 g. Find the percent yield.

Answer and explanation

88%

percent yield = (actual ÷ theoretical) × 100% = (22.0 ÷ 25.0) × 100% = 88%.

Problem 4

For N₂ + 3H₂ → 2NH₃, you start with 3.0 mol of N₂ and 6.0 mol of H₂. Which reactant is limiting, and how many moles of NH₃ can form?

Answer and explanation

H₂ is limiting; 4.0 mol of NH₃ forms.

Let each reactant predict the product. From N₂: 3.0 mol × (2 ÷ 1) = 6.0 mol NH₃. From H₂: 6.0 mol × (2 ÷ 3) = 4.0 mol NH₃. The smaller prediction (4.0 mol, from H₂) wins, so H₂ is limiting and 4.0 mol of NH₃ forms; some N₂ is left over.

Problem 5

In the Haber process N₂ + 3H₂ → 2NH₃, 28.02 g of nitrogen reacts with excess hydrogen. (a) What is the theoretical yield of ammonia, in grams? (b) If only 24.0 g of NH₃ is actually collected, what is the percent yield? (N₂ = 28.02 g/mol, NH₃ = 17.04 g/mol.)

Answer and explanation

(a) 34.08 g NH₃; (b) about 70.4%

(a) Moles of N₂ = 28.02 g ÷ 28.02 g/mol = 1.00 mol. Mole ratio NH₃ : N₂ = 2 : 1, so 2.00 mol NH₃ form, with mass 2.00 mol × 17.04 g/mol = 34.08 g (the theoretical yield). (b) percent yield = (24.0 ÷ 34.08) × 100% ≈ 70.4%.

Problem 6

2H₂+O₂→2H₂O: what does 3 mol O₂ require?

Answer and explanation

It requires 6 mol H₂ and can produce 6 mol H₂O if reactants are sufficient.

Response criteria

State a scientifically accurate conclusion.
Show the calculation or mechanism that supports it.
Include units, conditions, or limits where relevant.

Problem 7

Chapter review: For 2Mg+O₂→2MgO, use Mg=24.3 and MgO=40.3 g/mol. Find theoretical MgO from 4.86 g pure Mg with excess oxygen. Explain the mass increase.

Answer and explanation

4.86/24.3=0.200 mol Mg. The 1:1 ratio gives 0.200 mol MgO and 0.200×40.3=8.06 g MgO. The extra mass comes from incorporated oxygen; the complete mass account includes it.

Response criteria

Use the stated measurements or conditions, with correct units.
Connect the result to the chemical mechanism or evidence relationship.
Identify the assumption, model limit or further evidence requested.

Problem 8

N₂+3H₂→2NH₃ with 2molN₂ and 3molH₂: predict product.

Answer and explanation

H₂ limits; it supports 2molNH₃ and consumes 1molN₂, leaving 1molN₂. Compare potential products, not raw amounts.

Response criteria

State a scientifically accurate conclusion.
Show the calculation or mechanism that supports it.
Include units, conditions, or limits where relevant.

Problem 9

Chapter review: For N₂+3H₂→2NH₃, begin with 1.50 mol N₂ and 3.00 mol H₂. Find limiting reactant, theoretical NH₃, remaining N₂ and yield if 1.60 mol NH₃ is isolated.

Answer and explanation

H₂ limits. It uses 1.00 mol N₂, forms 2.00 mol NH₃ and leaves 0.50 mol N₂ in the complete model. Isolated yield is 1.60/2.00×100=80.0%; that number alone does not identify the cause of lost target.

Response criteria

Use the stated measurements or conditions, with correct units.
Connect the result to the chemical mechanism or evidence relationship.
Identify the assumption, model limit or further evidence requested.

Problem 10

How many mL of .100 M NaOH neutralize .002 mol HCl?

Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

20 mL

.002/.100 = .020 L =20 mL.

Problem 11

Compare dilution and neutralization equations.

Answer and explanation

Dilution preserves solute moles; neutralization transforms species according to a balanced equation.

Response criteria

State a scientifically accurate conclusion.
Show the calculation or mechanism that supports it.
Include units, conditions, or limits where relevant.

Problem 12

Chapter review: Mix 10.0 mL of 0.200 M AgNO₃ with 15.0 mL of 0.100 M NaCl. Assume complete AgCl precipitation and additive volumes. Find limiting ion, AgCl moles and remaining Ag⁺ concentration.

Answer and explanation

Ag⁺ starts at 0.00200 mol and Cl⁻ at 0.00150 mol, so chloride limits. AgCl forms 0.00150 mol; remaining Ag⁺ is 0.00050 mol in 0.0250 L, or 0.0200 M under the stated model.

Response criteria

Use the stated measurements or conditions, with correct units.
Connect the result to the chemical mechanism or evidence relationship.
Identify the assumption, model limit or further evidence requested.

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