Given : $H_{2} O_{2} + {KMnO}_{4} + H_{2} {SO}_{4} \to {MnSO}_{4} + H_{2} O + O_{2} + K_{2} {SO}_{4}$
which is/are correct statement(s)?

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Number of moles of ${MnSO}_{4}$ and $O_{2}$ produced in the ratio of $5 : 2$

Equivalent mass of $H_{2} {SO}_{4}$ is 29.4

Number of gm equivalent of $H_{2} O_{2}$ and ${KMnO}_{4}$ reacts in the ratio of $5 : 2$

$100 ml 2.27 {V H}_{2} O_{2}$ reacts completely with $160 ml 0.5 {M KMnO}_{4}$ in given reaction

answer is C.

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Detailed Solution

$5 \times (H_{2} O_{2} \to 2 H^{+} + O_{2} + 2 e^{-})$
$2 \times ({MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} + {Mn}^{2 +} + 4 H_{2} O)$
$5 H_{2} O_{2} + 2 {KMnO}_{4} + 3 H_{2} {SO}_{4} \to 5 O_{2} + 2 {MnSO}_{4} + 8 H_{2} O + K_{2} {SO}_{4}$
(A) No. of gm equivalent of $H_{2} O_{2} =$ no. of gm equivalent of ${KMnO}_{4}$
(B) $\frac{{Moles of MnSO}_{4}}{{moles of O}_{2}} = \frac{2}{5}$
(C) $E = \frac{98}{10} \times 3 = 29.4$
(D) $x_{H_{2} O_{2}} = x_{{KMnO}_{4}} \frac{2 \times 100 \times 0.2}{1000} = \frac{5 \times 0.5 V}{1000} \Rightarrow V = 16 ml$