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Q. A $0.518\, g$ sample of limestone is dissolved in $HCl$ and then the calcium is precipitated as $CaC _{2} O _{4}$. After filtering and washing the precipitate, it requires $40.0 mL$ of $0.250$ $N KMnO _{4}$ solution acidified with $H _{2} SO _{4}$ to titrate it as. The percentage of $CaO$ in the sample is:
$MnO _{4}^{-}+ H ^{+}+ C _{2} O _{4}^{2-} \longrightarrow Mn ^{2+}+ CO _{2}+2 H _{2} O$

Redox Reactions

Solution:

M eq. of $KMnO _{4}= M$ eq. of $C _{2} O _{4}^{-2}= M$ eq. of $CaCO _{3}$

$40 \times .25= M$ eq. of $CaO$

$\frac{10 \times 10^{-3}}{2}=$ Mole of $CaO$

$\% CaO =\frac{5 \times 10^{-3} \times 56 \times 100}{.518}=54 \%$