Question

3H2(g)+N2(g)→2NH3(g)

1.07 g H2 is allowed to react with 9.92 g N2, producing 2.33 g NH3.

1. What is the theoretical yield for this reaction under the given conditions?

Express your answer numerically in grams.

2.

What is the percent yield for this reaction under the given conditions?

Express the percentage numerically.

Answer 1

1)

Theoretiocal yield = 6.014g

Explanation

3H2(g) + N2(g) - - - - - - > 2NH3(g)

Stoichiometrically, 3moles of H2 react with 1mole of N2

given moles of H2 = 1.07g/2.02g/mol = 0.5297

given moles of N2 = 9.92g/ 28.014g/mol = 0.3541

0.3541 moles of N2 require 3× 0.3541= 1.062 moles of H2 but available moles of H2 is 0.5297. So, H2 is limiting reagent

stoichiometrically, 3 moles of H2 produce 2moles of NH3

0.5297 moles of H2 produce (2/3)×0.5297mol = 0.3531 moles of NH3

molar mass of NH3 = 17.031g/mol

Theoretical yield = 0.3531mol ×17.031g/mol = 6.014g

2)

38.74%

Explanation

percent yield = (Actual yield /Theoreical yield) ×100

percent yield = (2.33g/6.014g)×100 = 38.74%