Question

A $0.518$ g sample of limestone is dissolved in $HCl$ and the the calcium is precipitated as $Ca{C}_2{O}_4$. After filtering and washing the precipitate, it requires $40.0$ mL of $0.250NKMn{O}_4$ solution acidified with $H_{2}S{O}_4$ to titrate it as. The percentage of $CaO$ in the sample is :

$Mn{O}_4^{-}+H^{+}+C_2{O}_4^{2-}⟶M{n}^{2+}+C{O}_2+2H_{2}O$

• A. $54.0\%$
• B. $27.1\%$
• C. $42\%$
• D. $84\%$

Answer 1

The correct option is A $54.0\%$

Milliequivalents of $KMn{O}_4=$ Milliequivalents of $C_2{O}_4^{2-}=$ Milliequivalents of $CaC{O}_3$
$40\times 0.25=$ Milliequivalents of $CaO$
$\frac{10\times {10}^{-3}}{2}=$ Moles of $CaO$
$\%CaO=\frac{5\times {10}^{-3}\times 56\times 100}{0.518}$ $=54\%$