# According to this reaction: 2Al+6HBr -> 2AlBr_3 + 3H_2 If 4 moles of Al react with 8 moles of HBr, how many moles of H_2 are formed?

$6$ $\text{moles}$ of ${H}_{2} \left(g\right)$ are formed.
$2 A l \left(s\right) + 6 H B r \left(a q\right) \rightarrow 2 A l B {r}_{3} \left(a q\right) + 3 {H}_{2} \left(g\right) \uparrow$
You have the chemical equation, it tells you EXPLICITLY, that 2 moles of aluminum metal give 3 moles of dihydrogen gas upon reaction. You started with 4 moles of aluminum, thus $\frac{3}{2}$ that quantity of dihydrogen are formed.