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Stoichiometric Calculations

10.0 g sample...

Question

$10.0$ g sample of $C u_{2} O$ is dissolved in $d i l . H_{2} S O_{4}$ where it undergoes disproportionation quantitatively. The solution is filtered off and $8.3$ g pure $K I$ crystals are added to clear filtrate in order to precipitate $C u I$ with the liberation of $I_{2}$ . The solution is again filtered and boiled till all the $I_{2}$ is removed. Now excess of an oxidising agent is added to the filtrate which liberates $I_{2}$ again. The liberated $I_{2}$ now requires $10$ mL of $1.0 M$ $N a_{2} S_{2} O_{3}$ solution. The percentage by mass of $C u_{2} O$ in the sample is :

27.6

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28.6

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28.5

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none of the above

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Solution

The correct option is B $28.6$ %
$C u_{2} O \to C u^{2 +} + C u^{0}$
The solution after dissolution of $C u_{2} O$ in $d i l . H_{2} S O_{4}$ contains $C u^{2 +}$ ions and $C u$ . $C u^{2 +}$ ions react with $K I$ to give $C u I_{2}$ which is converted to $C u I$ $($ or $C u_{2} I_{2})$ and $I_{2}$ .
$C u^{2 +} + 2 I^{⊝} \to C u I_{2} \to (C u_{2} I_{2}$ or $C u I) + \frac{1}{2} I_{2}$
millimoles of $K I$ taken $= \frac{8.3}{166} \times 10^{3} = 50 (M w$ of $K I = 166)$
Now, $K I$ left unused reacts with oxidising agent to liberate $I_{2}$ again.
$2 I^{⊝} O x i d i s i n g a g e n t - --------- \to I_{2} 2 S_{2} {O_{3}}^{2 -} - ---- \to S_{4} {O_{6}}^{2 -} + I^{⊝}$
millimoles of $K I$ left $=$ millimoles of $S_{2} {O_{3}}^{2 -}$ used
$(n = \frac{2}{2} = 1) = (n = \frac{2}{2} = 1)$ $= 10 \times 1.0 \times 1 (n$ -factor $) = 10$
Therefore, millimoles of $K I$ used for $C u^{2 +} = 50 - 10 = 40$ .
Therefore, milli moles of $C u_{2} O = 20$
$⎡ ⎢ ⎣ \begin{matrix} C {+ 1 u}_{2} \to 2 + 2 C u ∴ M o l a r r a t i o o f C u_{2} : + 2 C u = 1 : 2 \end{matrix} ⎤ ⎥ ⎦$
$\frac{W e i g h t}{M w} \times 10^{3} = 20$ $(M w$ of $C u_{2} O = 143)$
$\frac{W e i g h t}{143} \times 10^{3} = 20$
$∴ W_{C u_{2} O} = 2.86$
% of $C u_{2} O = \frac{2.86 \times 100}{10.0} = 28.6$ %

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ml of

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C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

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The liberated iodine is titrated with hypo (

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2 C r O_{4}^{2 -} + 2 H^{+} ⟶ C r_{2} O_{7}^{2 -} + H_{2} O

C r_{2} O_{7}^{2 -} + 6 I^{-} + 14 H^{+} ⟶ 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O

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30

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K I O_{3} + 2 K I + 6 H C l ⟶ 3 I C l + 3 K C l + 3 H_{2} O

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