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 g of ${\mathrm{KNO}}_3$ is:
(Given : Atomic masses of H, O, N and K are 1, 16, 14 and 39, respectively.)