Question

$4.9$ g of$K_{2}C{r}_2{O}_7$ is taken to prepare 0.1 L of the solution. 10 mL of this solution is further taken to oxidise $S{n}^{2+}$ ion into $S{n}^{4+}$ ion $S{n}^{4+}$ so produced is used in second reaction to prepare $F{e}^{3+}$ ion then the millimoles of $F{e}^{3+}$ ion formed will be (assume all other components are in sufficient amount) [Molar mass of $K_{2}C{r}_2{O}_7=294$ g ].

• A. $5$
• B. $20$
• C. $10$
• D. none of these

Answer 1

Answer: (C)

$K_{2}C{r}_2{O}_7+S{n}^{2+}\to S{n}^{4+}+C{r}^{3+}$
$S{n}^{4+}F{e}^{2+}\to F{e}^{3+}$
Meq. of $S{n}^{4+}=$ Meq of $K_{2}C{r}_2{O}_7$
Meq. of $S{n}^{4+}=$ Meq of $F{e}^{3+}$
Meq. of $F{e}^{3+}=$ meq of $K_{2}C{r}_2{O}_7$
$N_{K_{2}C{r}_2{O}_7}=\frac{4.9\times 6}{294\times 01}=1$
millimol $\times n-$factor $=1\times 10$
millimol $=10$