Question

The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia were produced. What is the percent yield of this reaction?

30.8%
61.8%
20.7%
41.5%
More information is needed to solve this problem.

Answer 1

Answer : The percent yield of the reaction is, 61.8 %

Solution : Given,

Mass of $H_2$ = 450 g

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $H_2$.

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{450g}{2g/mole}=225moles$

Now we have to calculate the moles of $NH_3$

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction, we conclude that

As,  3 mole of $H_2$ react to give 2 mole of $NH_3$

So, 225 moles of $H_2$ react to give $\frac{225}{3}\times 2=150$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(150moles)\times (17g/mole)=2550g$

Theoretical yield of $NH_3$ = 2550 g

Experimental yield of $NH_3$ = 1575 g

Now we have to calculate the percent yield of the reaction.

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{1575g}{2550g}\times 100=61.8\%$

Therefore, the percent yield of the reaction is, 61.8 %