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  3. Chemistry
  4. NH 3 is produced according to the following reaction: N 2( g )+3 H 2( g ) arrow 2 NH 3( g ) In an experiment 0.25 mol of NH 3 is formed when 0.5 mol of N 2 is reacted with 0.5 mol of H 2. What is % yield?

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Q. $NH _{3}$ is produced according to the following reaction:
$N _{2}( g )+3 H _{2}( g ) \rightarrow 2 NH _{3}( g )$
In an experiment $0.25 mol$ of $NH _{3}$ is formed when $0.5 mol$ of $N _{2}$ is reacted with $0.5 mol$ of $H _{2}$. What is $\%$ yield?

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Solution:

$\%$ yield $=\frac{\text { produced mol of } NH _{3} \times 100}{\text { max. possible produced mole of } NH _{3}}$
$=\frac{0.25}{0.5 \times \frac{2}{3}} \times 100=75 \%$