Consider the reaction
$$ 4 \mathrm{HNO}_3(l)+3 \mathrm{KCl}(\mathrm{~s}) \rightarrow \mathrm{Cl}_2(\mathrm{~g})+\mathrm{NOCl}(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{~g})+3 \mathrm{KNO}_3(\mathrm{~s}) $$
The amount of $\mathrm{HNO}_3$ required to produce 110.0 g of $\mathrm{KNO}_3$ is: (Given : Atomic masses of $\mathrm{H}, \mathrm{O}, \mathrm{N}$ and K are $1,16,14$ and 39 , respectively.)