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.)