Consider the reaction

$$4 \mathrm{HNO}_{3}(1)+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})$$

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.)

$$ \begin{array}{r} 4 \mathrm{HNO}_{3}(\mathrm{l})+3 \mathrm{KCl}(\mathrm{s}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+ \\ 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s}) \end{array} $$ <br/><br/>$\because 110 \mathrm{~g}$ of $\mathrm{KNO}_{3} \Rightarrow$ moles of $\mathrm{KNO}_{3}=\frac{110}{101}$ <br/><br/>$=1.089 \mathrm{~mol}$ <br/><br/>As, 4 mole of $\mathrm{HNO}_{3}$ produces $3 \mathrm{~mol}$ of $\mathrm{KNO}_{3}$. <br/><br/>Hence, the moles of $\mathrm{HNO}_{3}$ required to produce <br/><br/>$1.089$ moles of $\mathrm{KNO}_{3}=\frac{4}{3} \times 1.089=1.452 \mathrm{~mol}$ <br/><br/>Hence, mass of $\mathrm{HNO}_{3}$ required is $1.452 \times 63$ <br/><br/>$=91.5 \mathrm{~g} \text { }$