# What molar quantity of dihydrogen gas results from the oxidation of a 5.9*mol quantity of aluminum metal?

Jan 19, 2017

Approx. $18 \cdot g$...........

#### Explanation:

We need a stoichiometric equation:

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

Given that $5.9 \cdot m o l$ of metal reacted, there are $5.9 \cdot m o l \times \frac{3}{2}$ $=$ $8.85 \cdot m o l$ $\text{dihydrogen gas}$ evolved. And what mass does this represent?

$\text{Mass of dihydrogen}$ $=$ 5.9*molxx3/2xx2.016*g*mol^-1=??g