Question

When 38.0 g of N2 is reacted with H2 and 40.12 g of NH3 are produced, what is the percent yield?

Answer 1

Answer:

52.80 % is the percent yield of the reaction.

Explanation:

Mass of nitrogen gas = 38.0 g

Moles of nitrogen = $\frac{38.0g}{17 g/mol}=2.235 mol$

$3H_2+N_2\rightarrow 2NH_3$

According to reaction, 1 moles of nitrogen gas gives 2 moles of ammonia, then 2.235 moles of nitrogen  will give:

$\frac{2}{1}\times 2.235mol=4.470 mol$  ammonia

Mass of 4.470  moles of ammonia

= 4.470 mol × 17 g/mol = 75.99 g

Theoretical yield of ammonia = 217.8 g

Experimental yield of ammonia = 40.12 g

The percentage yield of reaction:

$=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{40.12 g}{75.99 g}\times 100=52.80\%$

52.80 % is the percent yield of the reaction.