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+}}\xrightarrow{{}}{{N}_{2}}O(g)+4F{{e}^{2+}} $ $ +4{{H}^{+}}+{{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. $ 0.0307\text{ }M $
• B. $ 0.0316M $
• C. $ 0.1431\text{ }M $
• 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+}}\xrightarrow{{}}{{N}_{2}}O(g)+4F{{e}^{2+}} $ $ +4{{H}^{+}}+{{H}_{2}}O $ $ 6F{{e}^{2+}}+C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}\xrightarrow{{}}6F{{e}^{3+}} $ $ +2C{{r}^{3}}+7{{H}_{2}}O $ $ \frac{\begin{align} & 1C{{r}_{2}}O_{7}^{2-}\equiv 6\,F{{e}^{2+}}\equiv 3N{{H}_{2}}OH \\ & 1mol\,C{{r}_{2}}O_{7}^{2-}\equiv 3\,moloN{{H}_{2}}OH \\ \end{align}}{=\frac{23.61\times 0.0217}{1000}mol\,C{{r}_{2}}O_{7}^{2-}} $ $ =\frac{3\times 23.61\times 0.0217}{1000}mol\,N{{H}_{2}}OH $ $ =1.537\times {{10}^{-3}}mol\,\,N{{H}_{2}}OH $ in $ 50mL $ solution Molar concentration of $ N{{H}_{2}}OH $ $ =\frac{1.537\times {{10}^{-3}}}{0.050}=0.0307M $