Question

For the reaction
Zn + 2HCl ! ZnCl2 + H2 ,
what is the maximum amount of ZnCl2 which
could be formed from 10.58 g of Zn and 10.9 g
of HCl?
Answer in units of g

Answer 1

Task:

For the reaction

$$\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_2 + \mathrm{H}_2,$$

what is the maximum amount of $\mathrm{ZnCl}_2$ which could be formed from $10.58\,\mathrm{g}$ of Zn and $10.9\,\mathrm{g}$ of HCl?

Solution:

From the periodic table of elements:

$$\mathrm{MW}(\mathrm{Zn}) = 65.39\,\mathrm{g/mol}$$ $$\mathrm{MW}(\mathrm{ZnCl}_2) = 136.4\,\mathrm{g/mol}$$

The number of moles of Zn is

$$n(\mathrm{Zn}) = m(\mathrm{Zn}) / \mathrm{MW}(\mathrm{Zn}) = 10.58 / 65.39 = 0.1618\,\mathrm{mol}$$

The number of moles of HCl is

$$n(\mathrm{HCl}) = m(\mathrm{HCl}) / \mathrm{MW}(\mathrm{HCl}) = 10.9 / 36.5 = 0.299\,\mathrm{mol}$$

According to the chemical equation the number of moles of Zn is half the number of moles of HCl.

$$n(\mathrm{Zn}) = n(\mathrm{HCl}) / 2 = 0.299 / 2 = 0.150\,\mathrm{mol} \quad \text{(we have } 0.1618\,\mathrm{mol} \text{ of Zn)}.$$

That means that HCl is the limiting reactant.

The number of moles of $\mathrm{ZnCl}_2$ in this reaction is

$$n(\mathrm{ZnCl}_2) = n(\mathrm{HCl}) / 2 = 0.299 / 2 = 0.150\,\mathrm{mol}$$

The mass of $\mathrm{ZnCl}_2$ is

$$m(\mathrm{ZnCl}_2) = n(\mathrm{ZnCl}_2) \cdot \mathrm{MW}(\mathrm{ZnCl}_2) = 0.150 \cdot 136.4 = 20.46\,\mathrm{g}$$

Answer: $m(\mathrm{ZnCl}_2) = 20.46\,\mathrm{g}$