Question

Consider the balanced equation.


N2 + 3H2 Right arrow. 2NH3


What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3?
Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
17.8%
33.3%
35.7%
53.4%

Answer 1

Moles H2 = 26.3 g ÷ 2.016 g/mol = 13.04 mol
Moles NH3 (theoretical) = (2/3)(13.04) = 8.693 mol
Mass NH3 (theoretical) = 8.693 mol × 17.031 g/mol = 148.1 g

Percent yield = (79.0 g ÷ 148.1 g) × 100% = 53.4%

Answer: 53.4%