# If 30.2 g of aluminum react with HCI to produce aluminum chloride and hydrogen gas, how many liters of hydrogen are produced at STP?

Sep 10, 2016

Under $40 \cdot L$ of dihydrogen gas are generated.

#### Explanation:

We must use a stoichiometric equation:

$A l \left(s\right) + 3 H C l \left(a q\right) \rightarrow A l C {l}_{3} \left(a q\right) + \frac{3}{2} {H}_{2} \left(g\right) \uparrow$

This equation tells us that $26.98 \cdot g$ aluminum metal gives us $33.6 \cdot L$ of dihydrogen gas. Why? Because $1$ $m o l$ of gas occupies $22.4 \cdot L$ at $\text{STP}$ and $1.5 \cdot m o l$ were generated.

$\text{Moles of aluminum}$ $=$ $\frac{30.2 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.12 \cdot m o l$.

Given the stoichiometry, $\frac{3}{2} \times 1.12 \cdot \cancel{m o l} \times 22.4 \cdot L \cdot \cancel{m o {l}^{-} 1}$ are generated.