Chemistry - Mole Concept Question with Solution | TestHub

$$\begin{gathered} {\mathrm{N}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{NH}}_{3\left(\mathrm{g}\right)} \\ 20 \mathrm{g} 5 \mathrm{g} \mathrm{Given} \mathrm{mass} \\ \frac{20}{28} \frac{5}{2} \mathrm{Number} \mathrm{of} \mathrm{moles} \end{gathered}$$

To find the limiting reagent we can divide given number of moles with Stoichiometric coefficients, whichever will be the lower will be limiting reagent.

${\mathrm{N}}_2\to \frac{{\displaystyle \frac{20}{28}}}{1}=\frac{20}{28} {\mathrm{H}}_2\to \frac{{\displaystyle \frac{5}{2}}}{3}=\frac{5}{6}$

$\therefore {\mathrm{N}}_2$ is the Limiting Reagent.

According to the reaction one mole of nitrogen gas gives two moles of ammonia.

$\therefore \mathrm{n}\left({\mathrm{NH}}_3\right)=2\times \mathrm{n}\left({\mathrm{N}}_2\right)=2\times \frac{20}{28}$

Number of moles of ammonia formed will be $=1.42$