Question

In the reaction:
$2Al\left(s\right)+6HCl\left(aq\right)\to 2A{l}^{3+}\left(aq\right)+6C{l}^{-}\left(aq\right)+3H_2\left(g\right)$ Calculate the volume of $H_2$ gas (L) produced at $STP$(atm) for every 3 mol of $HCl$ consumed.

• A. $33.6\text{L}$
• B. $44.8\text{L}$
• C. $67.2\text{L}$
• D. $22.4\text{L}$

Answer 1

The correct option is A $33.6\text{L}$

According to the reaction,
6 mol of $HCl$ is consumed to produce 3 mol of $H_2$ gas, i.e. $3\times 22.4\text{L}$ of $H_2$ gas
3 mol of $HCl$ will produce = $\frac{3\times 22.4}{6}\times 3\text{L}=33.6\text{L}$