Question

What volume of $O_2\left(g\right)$ measured at 1atm and 273 K will be formed by action of 100 mL of 0.5 N $KMn{O}_4$ on hydrogen peroxide in an acid solution ? The skeleton equation for the reaction is
($KMn{O}_4+H_{2}S{O}_4+H_2{O}_2\to K_{2}S{O}_4+MnS{O}_4+O_2+H_2)$

• A. 0.12 litre
• B. 0.28 litre
• C. 0.56 litre
• D. 1.12 litre

Answer 1

The correct option is B 0.28 litre

$2{KMnO}_4+H_2{SO}_4+H_2{O}_2\to K_2{SO}_4+2{MnSO}_4+O_2+2H_2$

According to law of equivalance$:\to$

Gram$-$ Equivalent of each reactants and products are equal therefore

$\therefore$ gram equivalent of $KMn{O}_4=\frac{100}{1000}\times 0.5\because (gm-eqv=N\times V)$

$=0.05$

$\therefore gm-eqv$ of $KMn{O}_4=gm-eqv$ of $O_2=\frac{wt}{eqwt}=0.05$

$\Rightarrow \frac{wt}{8}=0.05$

$\Rightarrow wt=0.4gm$

eq.wt of $O_2=\frac{32}{4}=8gm$

$O_2\to {20}^{2-}\left(2e^{-}\right)\to nfactor=4$

mole $=\frac{wtof{O}_2}{32}=\frac{0.4gm}{32gm}=\frac{10}{8\times 100}=\frac{1}{80}$ mole

$\therefore$ volume $=0.28lt$