What volume of O_{2(g)} measured at standard | vedclass

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(C) The balanced chemical equation is:$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 	o K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$According to the law of equivalence,the numbe

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$0.28$

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(C) The balanced chemical equation is:$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 	o K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$According to the law of equivalence,the number of equivalents of $KMnO_4$ reacting is equal to the number of equivalents of $O_2$ produced.Equivalents of $KMnO_4 = 	ext{Normality} 	imes 	ext{Volume (in } L) = 0.5 \ N 	imes 0.1 \ L = 0.05 \ eq$.Since the equivalents of $O_2$ produced must also be $0.05$,and the equivalent weight of $O_2$ in this reaction (where $O^{-2} 	o O^0$,$n$-factor $= 4$) is $\frac{32}{4} = 8 \ g/eq$:Mass of $O_2 = 0.05 \ eq 	imes 8 \ g/eq = 0.4 \ g$.At standard conditions $(STP)$,$32 \ g$ of $O_2$ occupies $22.4 \ L$.Volume of $0.4 \ g$ of $O_2 = \frac{22.4 \ L}{32 \ g} 	imes 0.4 \ g = 0.28 \ L$.

What volume of $O_{2(g)}$ measured at standard condition will be formed by the action of $100 \ mL$ of $0.5 \ N \ KMnO_4$ on hydrogen peroxide in an acidic solution? The skeleton equation for the reaction is:
$KMnO_4 + H_2SO_4 + H_2O_2 \to KHSO_4 + MnSO_4 + O_2 + H_2O$
(Answer in $L$)

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What volume of $O_{2(g)}$ measured at standard condition will be formed by the action of $100 \ mL$ of $0.5 \ N \ KMnO_4$ on hydrogen peroxide in an acidic solution? The skeleton equation for the reaction is:$KMnO_4 + H_2SO_4 + H_2O_2 \to KHSO_4 + MnSO_4 + O_2 + H_2O$(Answer in $L$)