Question

Find out the weight of ammonia formed when $10$ moles of $N_2$ reacts with $20$ moles of $H_2$.

Answer 1

$N_2+3H_2⟶2N{H}_3$ $N_2:H_2:N{H}_3$

$1$ $3$ $2$ $\to$ Mole ratio is $1:$ $3:$ $2$

$10$ $30$

Here $H_2$ is limiting as $10$ moles of $N_2$ can react with $30$ moles of $H_2$ .

Hence only $\frac{20}{3}=6.67$ moles of $N_2$ reacts to produce $2\times 6.67$ moles of $N{H}_3$

$=13.44$ moles of $N{H}_3$

$=13.44\times 17$

$=228.48g$ of $N{H}_3$ is produced.