Salt cake $(N a_{2} S O_{4})$ is prepared as follows:
$2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H C l$
How much $80 %$ pure salt cake could be produced form $100.0$ g of $90 %$ pure salt in the above reaction?

43.92

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68.62

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87.84

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136.25

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Solution

The correct option is D $136.25$ g
The molar masses of NaCl and $N a_{2} S O_{4}$ are 58.5 g/mol and 142 g/mol respectively.
100.0 g of 90% pure NaCl corresponds to $100.0 \times \frac{90}{100} \times \frac{1}{58.4} = 1.54$ moles.
The number of moles of $N a_{2} S O_{4}$ formed from 1.54 moles of NaCl are $\frac{1.54}{2} = 0.77$ moles.
The mass of $N a_{2} S O_{4}$ obtained is $0.77 \times 142 = 109.8$ g.
But the salt cake is 80% pure.
Hence, the mass of the salt cake that can be obtained is $109.8 \times \frac{100}{80} = 136.25$ g.

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