Question

For decolourisation of $1$ mole of $KMn{O}_4$, the moles of $H_2{O}_2$ required is:

• A. $1/2$
• B. $3/2$
• C. $5/2$
• D. $7/2$

Answer 1

The correct option is C $5/2$

$2KMn{O}_4+3H_{2}S{O}_4{\to }_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5\left[O\right][H_2{O}_2+O\to H_{2}O+O_2]\times 5$
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$2KMn{O}_4+5H_2{O}_2+3H_{2}S{O}_4\to K_{2}S{O}_4+2MnS{O}_4+8H_{2}O+5O_2$
$2mol$ of $KMn{O}_4$ oxidise $5mol$ of $H_2{O}_2$
$1mol$ of $KMn{O}_4$ will oxidise $\frac{5}{2}mol$ of $H_2{O}_2$