Question

Moles of $KHC _2 O _4$ (potassium acid oxalate) required to reduce $100 \,mL$ of $0.02 \,M \,KMnO _4$ in acidic medium $\left(\right.$ to $\left.Mn ^{2+}\right)$ is

• A. $0.002$
• B. $0.005$
• C. $0.001$
• D. $0.007$

Answer 1

Answer: (B)

$5 C _2 O _4^{2-}+2 MnO _4^{-} \rightarrow Mn ^{2+}+10 CO _2$
$2 \,mol$ of $MnO _4^{-} \equiv 5 \,mol $ of $C _2 O _4^{2-}\left( KHC _2 O _4\right)$
$\frac{100 \times 0.02}{1000} mol \,MnO _4^{-}=0.005 \,mol \,KHC _2 O _4$