Consider the reaction

$4{\mathrm{HNO}}_3\left(\mathrm{l}\right)+3\mathrm{KCl}\left(\mathrm{s}\right)\to {\mathrm{Cl}}_2\left(\mathrm{g}\right)+\mathrm{NOCl}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+3{\mathrm{KNO}}_3\left(\mathrm{s}\right)$

The amount of ${\mathrm{HNO}}_3$ required to produce$110.0 \mathrm{g}$ of ${\mathrm{KNO}}_3$ is

(Given : Atomic masses of $\mathrm{H},\mathrm{O},\mathrm{N}$ and $\mathrm{K}$ are $1,16,14$ and $39$, respectively.)

$4{\mathrm{HNO}}_3\left(\mathrm{l}\right)+3\mathrm{KCl}\left(\mathrm{s}\right)\to {\mathrm{Cl}}_2\left(\mathrm{g}\right)+\mathrm{NOCl}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+3{\mathrm{KNO}}_3\left(\mathrm{g}\right)$

$\begin{array}{}\\ \end{array}$$$\begin{gathered} {\mathrm{HNO}}_3 {\mathrm{KNO}}_3 \\ \mathrm{Weight} \mathrm{Xgm} 110\mathrm{gm} \\ \mathrm{moles} \frac{\mathrm{x}}{63} \frac{110}{101} \end{gathered}$$

As per the stoichiometric equation$4 \mathrm{moles} \mathrm{of} {\mathrm{HNO}}_3\to 3 \mathrm{moles} \mathrm{of} {\mathrm{KNO}}_3$

$1\to \frac{3}{4}$

$\frac{\mathrm{x}}{63}\to \frac{3}{4}\times \frac{\mathrm{x}}{63}=\frac{110}{101}$

$\mathrm{x}=\frac{110\times 63\times 4}{101\times 3}=91.5\mathrm{gm}$