Question

If $340g$ of a mixture of $N_2$ and $H_2$ in the correct ratio gave $20$% yield of ${NH}_3$, the mass produced would be:

• A. $16g$
• B. $17g$
• C. $20g$
• D. $68g$

Answer 1

The correct option is D $68g$

The molar masses of For the reaction $N_2,H_2\text{and}{NH}_3$ are 28 g/mol, 2 g/mol and 17 g/mol respectively.

For the reaction, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

1 mole of $N_2$ will react with 3 moles of $H_2$ as per reaction stoichiometry to give 2 moles of $N{H}_3$.

1 mole of $N_2=1\times 28=28$ g

3 moles of $H_2=3\times 2=6$ g

2 moles of $N{H}_3=2\times 17=34$ g

28 g of $N_2$ will react with 6 g of $H_2$ as per reaction stoichiometry to give 34 g of $N{H}_3$.

$28+6=34$ g of mixture of $N_2$ and $H_2$ will react as per reaction stoichiometry to give 34 g of $N{H}_3$. This is for 100 % yield.

For 20 % yield, $34$ g of mixture of $N_2$ and $H_2$ will react as per reaction stoichiometry to give $34\times \frac{20}{100}=6.8$ g of $N{H}_3$.

For 20 % yield, $340$ g of mixture of $N_2$ and $H_2$ will react as per reaction stoichiometry to give $6.8\times \frac{340}{34}=68$ g of $N{H}_3$.

Hence, option D is correct.