Question

Treatment of hydroxyl amine $(N{H}_{2}OH)$ with an excess of $Fe(III)$ results in the formation of $NgO$ and an equivalent amount of $Fe(II)$ . $2N{H}_{2}OH+4F{e}^{3+}\to N_{2}O(g)+4F{e}^{2+}$ $+4H^{+}+H_{2}O$ Calculate the molar concentration of an $N{H}_{2}OH$ solution if $Fe(II)$ produced from 50 mL solution required 23.61 mL of $0.0217M$ $K_{2}C{r}_2{O}_7$

• A. $0.0307M$
• B. $0.0316M$
• C. $0.1431M$
• D. $0.1448M$

Answer 1

Answer: (A)

$F{e}^{2+}$ obtained by reduction of $F{e}^{3+}$ by $N{H}_{2}OH$ is estimated by $C{r}_2{O}_7^{2-}$ . $2N{H}_{2}OH+4F{e}^{3+}\to N_{2}O(g)+4F{e}^{2+}$ $+4H^{+}+H_{2}O$ $6F{e}^{2+}+C{r}_2{O}_7^{2-}+14H^{+}\to 6F{e}^{3+}$ $+2C{r}^3+7H_{2}O$ $\frac{\begin{array}{rl}& 1C{r}_2{O}_7^{2-}\equiv 6F{e}^{2+}\equiv 3N{H}_{2}OH\\ & 1molC{r}_2{O}_7^{2-}\equiv 3moloN{H}_{2}OH\end{array}}{=\frac{23.61\times 0.0217}{1000}molC{r}_2{O}_7^{2-}}$ $=\frac{3\times 23.61\times 0.0217}{1000}molN{H}_{2}OH$ $=1.537\times {10}^{-3}molN{H}_{2}OH$ in $50mL$ solution Molar concentration of $N{H}_{2}OH$ $=\frac{1.537\times {10}^{-3}}{0.050}=0.0307M$