What volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75g of HCl?

Answer 1

This is a limiting reagent (LR) problem. You know that when quantities are given for both reactanats.
Mg + 2HCl --> H2 + MgCl2
mols Mg = g/atomic mass = 50/24.3 = estimated 2
mols HCl = g/molar mass = 75/36.5 = estimated 2
1 mol Mg, according to the equation, reacts with 2 mols HCl so 2 mols Mg will require 4 mols HCl. You don't have that much HCl; therefore, HCl is the LR. So estimated 2 mols HCl will produce estimated 1 mol H2 gas. 1 mol H2 gas @ STP occupies 22.4 L. You should go through all of the calculations (not estimated) to obtain a more accurate number.