Chemistry - Mole Concept Question with Solution | TestHub

The given reaction is
$\underset{\text{lime}\text{stone}}{{\text{CaCO}}_{\text{3}}}\stackrel{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}{\to }{\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\stackrel{{\text{KMnO}}_{\text{4}}}{\to }\text{decolourizes}$
The redox changes are :
${\left({\mathrm{C}}^{3+}\right)}_2\to 2{\mathrm{C}}^{4+}+2\mathrm{e}$
$\mathrm{M}{\mathrm{n}}^{7+}+5\mathrm{e}\to \mathrm{M}{\mathrm{n}}^{2+}$
$\text{Meq}\text{.}\text{of}\text{CaO}\text{=}\text{Meq}\text{.}\text{of}{\text{CaCO}}_{\text{3}}\text{=}\text{Meq}\text{.}\text{of}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$
$\text{=}\text{Meq}\text{.}\text{of}{\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{=}\text{Meq}\text{.}\text{of}{\text{KMnO}}_{\text{4}}$
Hence$\text{Meq}\text{.}\text{of}\text{CaO}\text{=}\text{Meq}\text{.}\text{of}{\text{KMnO}}_{\text{4}}$
$\left[\mathrm{w}/(56/2)\right]\times 1000=45\times 0.2$
So$\text{wt}\text{.}\text{of}\text{CaO}\text{=}\text{0}\text{.252}\text{g}$
Here % of$\mathrm{C}\mathrm{a}\mathrm{O}$in lime stone =$\left(\frac{0.252}{0.256}\right)\times 100=45\mathrm{\%}$