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Stoichiometric Calculations

Copper metal ...

Question

Copper metal can be prepared from copper ore in the following steps :
$I : C u_{2} S (s) + O_{2} (g) ⟶ 2 C u (s) + S O_{2} (g)$
$I I : C u S (s) + O_{2} (g) ⟶ C u (s) + S O_{2} (g)$
Suppose an ore sample contains $11.0 %$ impurity in addition to a mixture of $C u S$ and $C u_{2} S$ . Heating $100.0$ g of the mixture produces $75.4$ g of copper metal with a purity of $89.5 %$ . What is the weight per cent of $C u S$ in the ore?(Round-off to the nearest integer)

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Solution

$C u_{2} S 159 g ⟶ 2 C u 127 g$
$C u S 95.5 g ⟶ C u 63.5 g$
There is an $11.0 %$ impurity. Thus, the mixture has $89.0$ g of $C u_{2} S$ and $C u S$ .
Let the amount of $C u_{2} S = x g$
The amount of $C u S = (89.0 - x) g$
Thus, $C u$ obtained from $x$ g $C u_{2} S = [\frac{127 x}{159}] g$
and from $(89.0 - x) g C u S = \frac{63.5 (89.0 - x)}{95.5} g$
Total amount of Cu obtained = $\frac{127 x}{159} + \frac{63.5 (89.0 - x)}{95.5} = \frac{75.4 \times 89.5}{100} (g i v e n)$ $= 67.483 g$
This gives $x$ $= C u_{2} S = 62.45$
% $o f C u_{2} S = 62.45$ %
% $o f C u S = 27.55$ %
Thus, the weight per cent of $C u S$ in the ore is $27.55 %$ .

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