# When 3.22 moles of Al reacts with 4.96 moles of HBr, how many or moles H_2 are formed?

Jan 23, 2017

Approx. $1.7 \cdot m o l$ of $\text{dihydrogen}$ are evolved.

#### Explanation:

We need (i) a stoichiometrically balanced equation:

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

And thus we know that 1 equiv aluminum requires 3 equiv hydrogen bromide. But (ii) we do not have a stoichiometric hydrogen bromide in the starting conditions. Realistically, the only way we can do this is to ASSUME that all the hydrogen bromide oxidizes the aluminum quantitatively, and assume that there is no incomplete oxidation.

And thus $\frac{4.96}{3} \cdot m o l$ of aluminum metal are oxidized, i.e. $1.65 \cdot m o l$, and $\frac{3}{2} \times 1.65 \cdot m o l$ $\text{dihydrogen gas}$ are produced as per the equation.