We drop 17.3 grams of magnesium into

Question

435 mL of a 2 M HCl solution. What is
the maximum volume of dry hydrogen that
could be produced by this reaction at STP?

Answer 1

Mg + 2HCl ==> H2 + MgCl2
This is a limiting reagent problem. We know that because amounts are given for BOTH reactants.
mols Mg = grams/molar mass = ?
mols HCl = M x L = ?

Using the coefficients in the balanced equation, convert mols Mg to mols H2.
Do the same for mols HCl to mols H2.
It is likely that the two values obtained for mols H2 gas will not agree which means one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
For the smaller value, at STP 1 mol H2 occupies 22.4L; therefore, volume H2 = mols H2 x 22.4 L/mol = ? L.