Stoichiometry — Drills · Grade 11 Science
Chapter: Stoichiometry & YieldWhat these drills covers
By the end of Stoichiometry you should be able to:
- I can read the mole ratio between any two substances directly from the coefficients of a balanced equation.
- I can carry out a mass-to-mass calculation by converting mass to moles, applying the mole ratio, and converting back to mass.
6 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
- atoms per gram
- grams
- moles
- milliliters
Answer and explanation
moles
Coefficients give mole ratios, which is why grams must be converted to moles before they can be used.
Problem 2
Answer and explanation
6 mol
The mole ratio N₂ : H₂ is 1 : 3, so 2.0 mol N₂ × 3 = 6.0 mol H₂.
Problem 3
Answer and explanation
10 mol
The mole ratio O₂ : H₂O is 1 : 2, so 5.0 mol O₂ × 2 = 10.0 mol H₂O.
Problem 4
Answer and explanation
88.02 g
Ratio C : CO₂ = 1 : 1, so 2.00 mol C → 2.00 mol CO₂; 2.00 mol × 44.01 g/mol = 88.02 g.
Problem 5
Answer and explanation
The coefficients are a ratio of moles, not of grams, because equal masses of different substances contain different numbers of particles. Converting to moles first lets you correctly apply the mole ratio, then you can convert back to grams at the end.
Problem 6
Answer and explanation
34.08 g
28.02 g ÷ 28.02 g/mol = 1.00 mol N₂; ratio NH₃ : N₂ = 2 : 1 gives 2.00 mol NH₃; 2.00 mol × 17.04 g/mol = 34.08 g.