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

52.80 % is the percent yield of the reaction.

Step-by-step explanation:

Mass of nitrogen gas = 38.0 g

Moles of nitrogen =
`(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:

`(2)/(1)* 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}}* 100`

`=(40.12 g)/(75.99 g)* 100=52.80\%`

52.80 % is the percent yield of the reaction.