Question

What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3?

Answer 1

Answer:

$\boxed{\text{53.3 \%}}$

Explanation:

MM:           2.016      17.03

           N₂ + 3H₂ ⟶ 2NH3

m/g:            26.3  

1. Theoretical yield

(a) Moles of H₂

$\text{Moles of H {_2} } = \text{26.3 g H {_2} } \times \dfrac{\text{1 mol H {_2} }}{\text{2.016 g H {_2} }} = \text{13.05 mol H {_2} }$

(b) Moles of NH₃

$\text{Moles of NH {_3} } = \text{13.05 mol H {_2} } \times \dfrac{\text{2 mol NH {_3} }}{\text{3 mol H {_2} }} = \text{8.697 mol NH {_3} }$

(c) Theoretical yield of NH₃

$\text{Mass of NH {_3} } = \text{8.967 mol NH {_3} } \times \dfrac{\text{17.03 g NH {_3} }}{\text{1 mol NH {_3} }} = \text{148.1 g NH {_3} }$

(d) Percent yield

$\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \% = \dfrac{\text{79.0 g}}{\text{148.1 g}} \times 100 \% = \textbf{53.3 \%}\\\\\text{The percent yield is }\boxed{\textbf{53.3 \%}}$

Answer 2

Answer:

53.3%

Explanation: