Question

If a sample of 0.213 g of KCN is treated with an excess of HCl, calculate the amount of HCN formed, in grams.

Answer 1

Answer on Question #62710 - Chemistry - Inorganic Chemistry

Task:

If a sample of 0.213 g of KCN is treated with an excess of HCl, calculate the amount of HCN formed, in grams.

Solution:

The reaction of:

$$\mathrm{KCN} + \mathrm{HCl} = \mathrm{HCN} + \mathrm{KCl}$$ $$n(\mathrm{KCN}) = n(\mathrm{HCN});$$ $$\frac{m(\mathrm{KCN})}{M(\mathrm{KCN})} = \frac{m(\mathrm{HCN})}{M(\mathrm{HCN})};$$ $$\frac{0.213}{(39 + 12 + 14)} = \frac{m(\mathrm{HCN})}{(1 + 12 + 14)};$$ $$m(\mathrm{HCN}) = \frac{0.213 \times 27}{65} = 0.088\,(g).$$

Answer:

$$m(\mathrm{HCN}) = 0.088\,g.$$

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