Question

According to this balanced equation, how many moles of bromine are required to produce 30 mol of aluminum bromide? 2 Al(s) + 3 Br2(g) → 2 AlBr3(s)

Answer 1

$2Al(s) + 3Br_2(g)\rightarrow 2AlBr_3(s)$

Covert moles of $AlBr_3$ to moles of $Br_2$

$30 moles AlBr_3 \frac{3 mol Br_2}{2 mol AlBr_3}=45 mol Br_2$