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

First we need to find the limiting reactant here. You can really do that visually in this case. You know that you need 3 times as much HBr as aluminum, and we can see that we don't have nearly enough HBr in this case, so we know that it's the limiting reactant.

Now that we know the LR, we just use a mole-to-mole conversion to see how many moles of H2 would be produced if all HBr was used...\[4.96mol\ HBr*\frac{3mol\ H_2}{6mol\ HBr}=2.48mol\ H_2\]