Question

Salt cake $\left(N{a}_{2}S{O}_4\right)$ is prepared as follows:

$2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HCl$

How much salt cake could be produced from $100.0$ g of $90\%$ pure salt in the above reaction?

• A. $109.8$ g
• B. $54.9$ g
• C. $36.6$ g
• D. $209.8$ g

Answer 1

The correct option is A $109.8$ 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.