Question

$4.00$ moles of $N_2$ and $3.00$ moles of $H_2$ are allowed to react in the presence of iron as a catalyst according to the reaction:
$N_2\left(g\right)+3H_2\left(g\right)\stackrel{Fe}{\to }2N{H}_3\left(g\right).$ The number of molecules of the gas left unreacted will be

• A. $1.86\times {10}^{23}$
• B. $6.02\times {10}^{23}$
• C. $18\times {10}^{23}$
• D. $1.20\times {10}^{23}$

Answer 1

The correct option is C $18\times {10}^{23}$

$N_2+3H_2\to 2N{H}_3$
According to the balanced equation, one mole of $N_2$ reacts with $3$ moles of $H_2$
So, $4$ moles of $N_2$ will require $12$ moles of $H_2$ but only $3$ moles of $H_2$ is given. Hence $H_2$ is the limiting reagent.
$\therefore$ moles of $N_2$ left $=4-1=3$
$\therefore$ Number of molecules of $N_2$ left $=3\times 6\times {10}^{23}=18\times {10}^{23}$