Question

2Al+6HBr -> 2AlBr3+3H2
When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed?

Answer 1

This is a limiting reactant problem. First calculate how much H2 would be formed from the 3.22 moles of Al, then with the 4.96 moles of HBr. Whichever one produces less is the limiting reactant, and that means that no more than that amount can be made.
\[3.22mol\ Al*\frac{3mol\ H_2}{2mol\ Al}=4.83mol\ H_2\]\[4.96mol\ HBr*\frac{3mol\ H_2}{6mol\ HBr}=2.48mol\ H_2\]The limiting reactant here is the HBr, so 2.48 moles of hydrogen gas are formed.