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$\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
The standard reduction potentials are given as below $\left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)$
$\mathrm{E}_{\mathrm{MmO}_4^{-} / \mathrm{Mm}^{2+}}^{\circ}=+1.51 \mathrm{~V}$
$\mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}$
If the equilibrium constant of the above reaction is given as $K_{\text {eq }}=10^x$, then the value of $x=$____ (nearest integer)


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