Answer to Question #312195 in Chemistry for Whitney Gray

Question

PRACTICE 1

1. What is the percent yield if 5.50 grams of hydrogen where used in a reaction that only produced 20.4 grams NH₃?

3H² + N² □ 2 NH₃

Answer 1

Solution:

The molar mass of hydrogen gas (H₂) is 2.016 g/mol

Therefore,

Moles of H₂ = (5.50 g H₂) × (1 mol H₂ / 2.016 g H₂) = 2.728 mol H₂

Balanced chemical equation:

3H₂ + N₂ → 2NH₃

According to stoichiometry:

3 mol of H₂ produces 2 mol of NH₃

Thus, 2.728 mol of H₂ produces:

(2.728 mol H₂) × (2 mol NH₃ / 3 mol H₂) = 1.819 mol NH₃

The molar mass of NH₃ is 17.031 g/mol

Therefore,

Mass of NH₃ = (1.819 mol NH₃) × (17.031 g NH₃ / 1 mol NH₃) = 30.98 g NH₃

Percent yield = (Actual yield / Theoretical yield) × 100%

Actual yield of NH₃ is 20.4 grams

Theoretical yield of NH₃ is 30.98 grams

Therefore,

Percent yield of NH₃ = (20.4 g / 30.98 g) × 100% = 68.85%

Percent yield of NH₃ = 68.85%

Answer: The percent yield of NH₃ is 68.85%