Chemistry - Mole Concept Question with Solution | TestHub

$+4$ $-1$ 0

$Mn{O}_2+HCl⟶MnC{l}_2+2H_{2}O+C{l}_2...(i)$

$n=2$ $n=2$

$0-1$ $0-1$

$C{l}_2+KI⟶I_2+KCl...(ii)$

$n=2$ $n=2$

Therefore, Equivalent of $I_2$ = equivalent of $C{l}_2$ reacted in equation (ii),

= equivalent of $C{l}_2$ produced in equation (i),

= equivalent of $Mn{O}_2$ reacted in equation (i).

=$\frac{2\times 8.7}{87}$

$\therefore 2\times \frac{wt.of{I}_2}{Molecularwt.of{I}_2}=2\times \frac{8.7}{87}⟹weightof{I}_2=25.4g.$