How much $74.1 %$ pure $N a_{2} S O_{4}$ could be produced from $245$ g of $93.9 %$ pure $N a C l$ ?
$2 N a C l + H_{2} S O_{4} ⟶ N a_{2} S O_{4} + 2 H C l$

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Solution

$2 N a C l + H_{2} S O_{4} ⟶ N a_{2} S O_{4} + 2 H C l$
$2 m o l 1 m o l$
$2 \times 58.5 g$
$117$ g $N a C l$ yield = $142$ g pure $N a_{2} S O_{4}$
$[245 \times \frac{93.9}{100}]$ g $N a C l$ yield $= \frac{142 \times 245 \times 93.9}{117 \times 100}$ g pure $N a_{2} S O_{4}$
$= 279.21$ g pure $N a_{2} S O_{4}$
$= 279.21 \times \frac{100}{74.1}$ of $74.1$ % purity
$= 376.80 g$
Hence, $376.80$ g of $74.1 %$ pure $N a_{2} S O_{4}$ could be produced from $245$ g of $93.9 %$ pure $N a C l$ .

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