We drop 34.6 grams of magnesium into
705 mL of a 2 M HCl solution. What is
the maximum volume of dry hydrogen that
could be produced by this reaction at STP?
This is a limiting reagent (LR) problem and stoichiometry problem rolled into one.
mols HCl = M x L = 2 x 0.705 = 1.41
mols Mg = g/atomic mass = 34.6/24.3 = 1.42
Mg + 2HCl ==> MgCl2 + H2(g)
How many mols H2 can be obtained from 1.42 mol Mg? That's 1.42.
How many mols H2 can be obtained from 1.41 mols HCl? That's 1.41/2 = 0.705. In LR problems the small number always wins so you can get 0.705 mols H2 and at STP each mol will occupy 22.4 L.
Then mols x 22.4 L/mol = L H2.