# When 4.22 moles of Al reacts with 5.0 moles of HBr, how many moles of H_2 are formed?

Mar 26, 2016

$2.5 m o l$

#### Explanation:

First write a balanced chemical equation for the reaction of the metal with the acid :

$2 A l + 6 H B r \to 2 A l B {r}_{3} + 3 {H}_{2}$.

This represents the mole ratio in which the reactants combine and in which the products are formed.

Since $2$ moles $A l$ requires $6$ moles $H B r$, it implies that, given $4.22$ moles $A l$ and $5.0$ moles $H B r$, then $H B r$ is in excess and $A l$ is the limiting reagent.

Hence all $5.0 m o l H B r$ will react with $\frac{5.0}{3} = 1.667 m o l A l$ to form $1.667 m o l A l B {r}_{3}$ and $\frac{5.0}{2} = 2.5 m o l {H}_{2}$

Mar 26, 2016

$2.5$ $m o l$ dihydrogen gas are formed.
$A l \left(s\right) + 3 H B r \left(a q\right) \rightarrow A l B {r}_{3} \left(a q\right) + \frac{3}{2} {H}_{2} \left(g\right) \uparrow$
Clearly, hydrobromic acid is in deficiency (complete rxn requires $> 12 \cdot m o l$ $H B r$). From the equation, $\frac{1}{2}$ $m o l$ dihydrogen gas are evolved per mole of $H B r$. Since $5$ $m o l$ $H B r$ are used, half of this quantity of ${H}_{2}$ evolve.