Question

The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 3H2(g)+N2(g)→2NH3(g)3H2(g)+N2(g)→2NH3(g) The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation.1.29g H2 is allowed to react with 9.55g N2, producing 1.49g NH3.
What is the theoretical yield for this reaction under the given conditions?
What is the percent yield for this reaction under the given conditions?

Answer 1

Answer: a. 7.31 g

b. 20.4 %

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} H_2=\frac{1.29g}{2g/mol}=0.645moles$

$\text{Moles of} N_2=\frac{9.55}{28g/mol}=0.341moles$

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

According to stoichiometry :

3 moles of $H_2$ require = 1 mole of $N_2$

Thus 0.645 moles of $H_2$ will require=$\frac{1}{3}\times 0.645=0.215moles$  of $N_2$

Thus $H_2$ is the limiting reagent as it limits the formation of product and $N_2$ is the excess reagent.

As 3 moles of $H_2$ give = 2 moles of $NH_3$

Thus 0.645 moles of $H_2$ give =$\frac{2}{3}\times 0.645=0.430moles$  of $NH_3$

Mass of $NH_3=moles\times {\text {Molar mass}}=0.430moles\times 17g/mol=7.31g$

Thus theoretical yield for this reaction under the given conditions is 7.31 g.

b) ${\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%$

${\text {percentage yield}}=\frac{1.49g}{7.31g}\times 100\%=20.4\%$

The percent yield for this reaction under the given conditions is 20.4 %