Consider the following equations :

$2{\mathrm{F}\mathrm{e}}^{2+}+{\mathrm{H}}_2{\mathrm{O}}_2\to \mathrm{x}\mathrm{A}+\mathrm{y}\mathrm{B}$ (in basic medium)

(in acidic medium)

$2{\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{x}}^{\mathrm{\text{'}}}\mathrm{C}+{\mathrm{y}}^{\mathrm{\text{'}}}\mathrm{D}+{\mathrm{z}}^{\mathrm{\text{'}}}\mathrm{E}$

(in acidic medium)

The sum of the stoichiometric coefficients $x,y,x^{\mathrm{\text{'}}},y^{\mathrm{\text{'}}}$ and $z^{\mathrm{\text{'}}}$ for products $A,B,C,D$ and $E$ , respectively, is

 

At $20.0\mathrm{m}\mathrm{L}$ solution containing $0.2\mathrm{g}$ impure ${\mathrm{H}}_2{\mathrm{O}}_2$ reacts completely with $0.316\mathrm{g}$ of ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4$ in acid solution. The purity of ${\mathrm{H}}_2{\mathrm{O}}_2$ (in %) is (mol. wt. of ${\mathrm{H}}_2{\mathrm{O}}_2=34$ ; mol. wt. of ${\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4=158$ ) 

On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid $\left(C\right)$ is: