Chemistry - Stoichiometry Question with Solution | TestHub

$${\left({\mathrm{Cu}}^{+}\right)}_2\to {\mathrm{Cu}}^{2+}+\mathrm{Cu}$$

The solution after dissolution of ${\mathrm{Cu}}_2\mathrm{O}$ in dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$ contains ${\mathrm{Cu}}^{2+}$ ions react with KI to give ${\mathrm{CuI}}_2$ which is converted to CuI and ${\mathrm{I}}_2$.

${\mathrm{Cu}}^{2+}+2\mathrm{I}\to {\mathrm{CuI}}_2+\frac{1}{2}{\mathrm{I}}_2$

Millimole of KI taken $=\frac{8.3}{166}\times 1000=50$

Now, $KI$ left unused reacts with oxidizing agent to liberate $I_2$ again. $2\mathrm{I}\stackrel{\text{ Oxidant }}{\to }{\mathrm{I}}_2\stackrel{2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3}{\to }{\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6+2\mathrm{NaI}$

∴ Millimole of KI left $=$ Millimole of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ used (mole ratio of $I^{-}$to ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ is $1:1)=10\times 1.0=20$

Therefore, millimole of $KI$ used for $C{u}^{2+}=50-10=40$

∴ Millimole of ${\mathrm{Cu}}_2\mathrm{O}=20$

$\left[\begin{array}{c}{\mathrm{Cu}}_2^{+}\to 2{\mathrm{Cu}}^{2+}\therefore \text{ Mole ratio }{\mathrm{Cu}}_2^{+}:{\mathrm{Cu}}_2^{+}::1:2\end{array}\right]$

or $\frac{W}{M}\times 1000=20$ $\frac{W\times 1000}{143}=20;\therefore W_{{\mathrm{Cu}}_2\mathrm{O}}=2.86$

∴ % of ${\mathrm{Cu}}_2\mathrm{O}=\frac{2.86\times 100}{3}=95.33$