Question

What is the approximate volume (in $mL$ ) of $10 \,vol\, H _{2} O _{2}$ solution that will react completely with $1\, L$ of $0.02\, M\, KMnO _{4}$ solution in acidic medium?

• A. 56.05
• B. 113.5
• C. 90.8
• D. 75.75

Answer 1

Answer: (A)

The balanced equation between the reaction of $KMnO _{4}$ and $H _{2} O _{2}$ in acidic medium is as follows:

$2 KMnO _{4}+5 H _{2} O _{2}+3 H _{2} SO _{4} \longrightarrow 5 O _{2}+2 MnSO _{4}+ K _{2} SO _{4}+8 H _{2} O$

Thus, $2$ moles of $KMnO _{4} \equiv 5$ moles of $H _{2} O _{2} \equiv 5$ moles of $O _{2}(g)$ at $STP$.

According to above relation:

$34 \,g$ of $H _{2} O _{2}$ will give $=11.2\, L$ of $O _{2}$ at $STP$.

$\therefore (34 \times 5)$, i.e. $170 g$ of $H _{2} O _{2}$ will give $=\frac{11.2 \times 170}{34}$

$=56\, L$

$\because 1\, L$ of $H _{2} O _{2}$ give $10 \,L$ of $O _{2}$ at $STP$.

$\therefore 1\, mL$ of $H _{2} O _{2}$ give $10 mL$ of $O _{2}$ at $STP$.

and $56\, mL$ of $H _{2} O _{2}$ will give $560\, mL$ of $O _{2}$

$\therefore $ Volume of $H _{2} O _{2}=56\, mL$