# In the reaction 2Al +6HBr -> 2AlBr_3 + 3H_2, when 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H_2 are formed?

Feb 5, 2016

$2.48 m o l$
Since 2 moles aluminuim are required for every 6 moles of hydrogen bromide, yet we only have 3.22 moles aluminium and 4.96 moles hydrogen bromide, it implies that $H B r$ will get used up first and is hence the limiting reactant, while $A l$ is in excess.
So all $4.96 m o l H B r$ is used up and reacts with $\frac{4.96}{3} = 1.653 m o l A l$ to produce $\frac{4.96}{2} = 2.48 m o l {H}_{2}$.