1. Write the balanced chemical equation:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

2. Determine the limiting reactant:

* Moles of Mg:

* Molar mass of Mg = 24.31 g/mol

* Moles of Mg = (50.0 g) / (24.31 g/mol) = 2.06 mol

* Moles of HCl:

* Assume "equivalent" means equal moles of HCl as Mg.

* Moles of HCl = 2.06 mol

* Stoichiometry: The balanced equation shows a 1:1 mole ratio between Mg and H₂. Since we have equal moles of Mg and HCl, Mg is the limiting reactant because it will be consumed first.

3. Calculate moles of H₂ produced:

* Using the mole ratio from the balanced equation, 1 mol Mg produces 1 mol H₂.

* Moles of H₂ = 2.06 mol Mg × (1 mol H₂ / 1 mol Mg) = 2.06 mol H₂

4. Calculate the volume of H₂ at STP:

* At STP (Standard Temperature and Pressure: 0°C and 1 atm), 1 mole of any ideal gas occupies 22.4 L.

* Volume of H₂ = 2.06 mol H₂ × (22.4 L/mol) = 46.1 L

Therefore, 46.1 L of hydrogen gas is produced at STP from the reaction of 50.0 g Mg and equivalent 75 HCl.