Question

Zinc and hydrochloric acid react according to the reaction $\mathrm{Zn}\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)\to \mathrm{ZnCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$. If $0.40\mathrm{mol}\mathrm{Zn}$ are added to hydrochloric acid containing $0.62\mathrm{mol}\mathrm{HCl}$, how many moles of ${\mathrm{H}}_2$ are produced?

Answer 1

Step 1: Write the equation specifying no. of moles involved

$$\begin{gathered} \mathrm{Zn}\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)\to \mathrm{ZnCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right) \\ \mathrm{No}.\mathrm{of}\mathrm{moles}1211 \end{gathered}$$

Step 2 : Finding limiting reagent

• $1$ mole of $\mathrm{Zn}$ reacts with $2$ moles of $\mathrm{HCl}$ . So $0.4$ moles of $\mathrm{Zn}$ react with $0.8$ moles of $\mathrm{HCl}$.
• But $0.8$ moles of $\mathrm{HCl}$ is not available.
• So $\mathrm{HCl}$ is a limiting reagent.
• Hence, $\mathrm{HCl}$ will decide the quantity.

Step 3: Finding moles of ${\mathrm{H}}_2$ produced

• $2$ moles of $\mathrm{HCl}$ reacts with $1$ mole of $\mathrm{Zn}$ .
• So $0.62$ moles of $\mathrm{HCl}$ reacts with $=\frac{1}{2}\times 0.62=0.31\mathrm{moles}\mathrm{of}\mathrm{Zn}$ .
• If a mole of $\mathrm{Zn}$ gives $1$ mole of ${\mathrm{H}}_2$ .
• Then $0.31$ moles of $\mathrm{Zn}$ will give $0.31$ moles of ${\mathrm{H}}_2$

Therefore, the moles Of ${\mathrm{H}}_2$ produced is $0.31\mathrm{moles}$.