Question

A sample of $MnS{O}_4\cdot 4H_{2}O$ is strongly heated in air. The residue $\left(M{n}_3{O}_4\right)$ left was dissolved in $100mL$ of $0.1NFeS{O}_4$ containing $dil.H_{2}S{O}_4$. This solution was completely reacted with $50mL$ of $KMn{O}_4$ solution. $25mL$ of this $KMn{O}_4$ solution was completely reduced by $30mL$ of $0.1NFeS{O}_4$ solution. Calculate the mass of $MnS{O}_4\cdot 4H_{2}O$ in sample. (in g as nearest integer)

Answer 1

$MnS{O}_4\cdot 4H_{2}O\stackrel{\Delta }{\to }M{n}_3{O}_4$
The residue $M{n}_3{O}_4$ is dissolved in $FeS{O}_4$, which is reduced from $M{n}^{\frac{8}{3}+}$ to $M{n}^{2+}$,
i.e., $(M{n}^{\frac{8}{3}+}{)}_3+2e⟶3M{n}^{2+}$ $[\because E_{M{n}^{2+}}=\frac{M}{\frac{2}{3}}]$
For the reaction of $KMn{O}_4$ with $FeS{O}_4$,
$Meq.ofKMn{O}_4=Meq.ofFeS{O}_4$
$25\times N=30\times 0.1$
$\therefore N_{KMn{O}_4}=\frac{3}{25}$
Meq. of $FeS{O}_4$ added to $M{n}_3{O}_4=100\times 0.1=10$
Meq. of $FeS{O}_4$ left after reaction with $M{n}_3{O}_4=Meq.ofKMn{O}_{4}used=50\times \frac{3}{25}=6$
$\therefore$ Meq. of $FeS{O}_4$ used for $M{n}_3{O}_4=10-6=4$
$\therefore$ Meq. of $MnS{O}_4=$ Meq. of $M{n}_3{O}_4=4$
$\because$ As reported above $E_{MnS{O}_4\cdot 4H_{2}O}=\frac{3M}{2}=\frac{223\times 3}{2}$
$\therefore \frac{w_{MnS{O}_4}}{\frac{3M}{2}}\times 1000=4$
$\therefore w_{MnS{O}_4}=\frac{4\times 3\times 223}{2\times 1000}=1.338g$