Chemistry - Mole Concept Question with Solution | TestHub

The reaction of KHC₂O₄ and KMnO₄ is:

$KH{C}_2{O}_4+Mn{O}_4^{-}\stackrel{H^{+}}{\to }C{O}_2+M{n}^{+2}+H_{2}O$

n=2 n=5

As the MnO₄⁻ is the oxidizing agent here.

So, equivalents of KHC₂O₄ = equivalents of MnO₄⁻.

$\frac{a}{Eq.wt.ofKH{C}_2{O}_4}0.1\times 0.02\times 5\dots (1)$

Now, the reaction of KHC₂O₄ with Ca(OH)₂ is a neutralization reaction:

So, no. of equivalents of KHC₂O₄ = no. of equivalents of Ca(OH)₂.

$\frac{b}{Eq.wt.ofKH{C}_2{O}_4}=eq.ofCa(OH{)}_2$

$\frac{b}{Eq.wt.ofKH{C}_2{O}_4}=0.1\times 0.05\times 2\dots (2)$

If the molecular weight of KHC₂O₄ is M, then in equation (1),

Equivalent weight of KHC₂O₄ = $M/2$,

and in equation (2), the equivalent weight of KHC₂O₄ = $M/1$.

By dividing (1) by (2), we get:

$\frac{\frac{a}{\frac{M}{2}}}{\frac{b}{\frac{M}{1}}}=\frac{0.1\times 0.02\times 5}{0.1\times 0.05\times 2}$

Therefore, $2a=b$.