Stoichiometry — Practice · Grade 11 Science

Chapter: Stoichiometry & Yield

What this practice covers

By the end of Stoichiometry you should be able to:

9 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: stoichiometry, mole ratio, mass-to-mass calculation, given, unknown.

Problem 1

In the balanced equation N₂ + 3H₂ → 2NH₃, what is the mole ratio of H₂ to NH₃?
  1. 3 : 1
  2. 3 : 2
  3. 1 : 1
  4. 2 : 3
Answer and explanation

3 : 2

The coefficients give H₂ : NH₃ = 3 : 2, read straight from the balanced equation.

Problem 2

For 2H₂ + O₂ → 2H₂O, how many moles of water form from 4.0 mol of hydrogen reacting with excess oxygen?
Answer and explanation

4 mol

The mole ratio of H₂O to H₂ is 2 : 2 = 1 : 1, so 4.0 mol H₂ → 4.0 mol H₂O.

Problem 3

For N₂ + 3H₂ → 2NH₃, how many moles of ammonia form from 9.0 mol of hydrogen (with excess nitrogen)?
Answer and explanation

6 mol

9.0 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂) = 6.0 mol NH₃.

Problem 4

For C + O₂ → CO₂, what mass of carbon dioxide is produced when 1.00 mol of carbon burns completely? (Molar mass CO₂ = 44.01 g/mol.)
Answer and explanation

44.01 g

The mole ratio of CO₂ to C is 1 : 1, so 1.00 mol C → 1.00 mol CO₂, and 1.00 mol × 44.01 g/mol = 44.01 g.

Problem 5

For 2H₂ + O₂ → 2H₂O, what mass of water forms from 4.04 g of hydrogen reacting with excess oxygen? (Molar masses: H₂ = 2.02 g/mol, H₂O = 18.02 g/mol.)
Answer and explanation

36.04 g

4.04 g ÷ 2.02 g/mol = 2.00 mol H₂; ratio H₂O:H₂ = 1:1 gives 2.00 mol H₂O; 2.00 mol × 18.02 g/mol = 36.04 g.

Problem 6

For N₂+3H₂→2NH₃, find NH₃ from 1.5 mol N₂ with excess H₂. Show the reasoning, not just the conclusion.
Hints

Mole ratios come from coefficients, not molar masses.

Answer and explanation

1.5×2/1=3.0 mol NH₃ theoretical yield.

Problem 7

Why must a grams-to-grams problem usually pass through moles?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Balanced coefficients relate particle counts/moles, so masses must be converted using each substance's molar mass.

Problem 8

Theoretical product is conditional on purity, partners and reaction extentTheoretical product is conditional on purity, partners and reaction extentConditions behind a theoretical amountCorrect balanced reaction and molar massesStarting purity and available reacting amountSufficient partner / stated completeness / no competing pathA calculation predicts under assumptions; it does not measure yield.
Original assumption organizer, not a complete kinetic/equilibrium calculation or a general promise of quantitative conversion.
Interpret the supplied visual. What changes if only 80% of a 10.0 g starting sample is the specified reactant?
Answer and explanation

Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.

A theoretical yield is the product amount predicted by a stated chemical model and available reacting quantities. An impure sample contains less target reactant than its total mass, an insufficient partner can limit conversion, and an equilibrium or competing pathway can reduce the modeled target formation. These are distinct limitations, not reasons to abandon conservation. State the reaction, purity, available amounts and completeness assumptions before presenting a number as an experimental outcome.

Only 8.00 g is available for that reactant’s stoichiometric conversion under the stated mass-fraction model.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 9

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.

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.

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