Question

For decolourisation of 1 mol 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 $5/2$

Eq of $Mn{O}_4^{\ominus }\equiv$ Eq of $H_2{O}_2$
$(n=5)$ $(n=2)$

$\frac{1}{5}$ mol $\equiv \frac{1}{2}$ mol

1 mol of $Mn{O}_4^{\ominus }=\frac{5}{2}$ mol of $H_2{O}_2$