Question

The percentage of $Mn{O}_2$ in a mineral specimen is $y\times {10}^{-1}$. If the iodine liberated by a $0.1344$ g sample in the net reaction, $Mn{O}_2\left(s\right)+4H^{+}+2I^{-}\to M{n}^{2+}+I_2+2H_{2}O$, require $32.30$ mL of $0.07220M$ $N{a}_2{S}_2{O}_3$. Then, $y$ is $(Mn{O}_2=87gmo{l}^{-1})$ :

Answer 1

The reactions are as follows:
$Mn{O}_2\left(s\right)+4H^{+}+2I^{-}\to M{n}^{2+}+I_2+2H_{2}O$
$I_2+2{S_2{O}_3}^{2-}\to {S_4{O}_6}^{2-}+2I^{-}$
Equivalent of $Mn{O}_2$ = Equivalents of ${S_2{O}_3}^{2-}$
Equivalent of $Mn{O}_2$ = $43.5$
$0.07220M$ ${S_2{O}_3}^{2-}$ = $0.07220N$ ${S_2{O}_3}^{2-}$
Equivalents of ${S_2{O}_3}^{2-}$=$\frac{0.07220\times 3230}{1000}=2.332\times {10}^{-3}$
Equivalents of $Mn{O}_2$ = $2.332\times {10}^{-3}$
Pure $Mn{O}_2$ in the sample = $2.332\times {10}^{-3}\times 43.5g$
$\%$ of pure $Mn{O}_2=\frac{0.1014}{0.1344}\times 100$ = $75.48\%$ $=755\times {10}^{-1}\%$.