5.2 grams of magnesium metal are place in 400.0 mL of 1.0 M hydrochloric acid. The balanced equation for the reaction is:

Mg + 2HCl = MgCl2 + H2. How many moles of H2 gas were produced?

mols Mg = g/atomic mass = 5.2/24.3 = about 0.21 but you need a more accurate figure.

mols HCl = L x M = 0.4 x 1M = about 0.4

Convert mols Mg to mols H2 = about 2 x (1 mol H2/1 mol Mg) = about 0.21 mol H2
Convert mols HCl to mols H2. That's 0.4 x (1 mol H2/2 mols HCl) = about 0.2 mol H2
You see the values don't agree for mols H2 produced. This is a limiting reagent problem and the correct value is ALWAYS the smaller value; therefore, 0.2 mol will be H2 produced and HCl is the limiting reagent.

To find the number of moles of H2 gas produced in the reaction, we need to use the balanced equation to determine the stoichiometry of the reaction.

The balanced equation is:
Mg + 2HCl = MgCl2 + H2

From the equation, we can see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2 gas.

First, let's calculate the number of moles of Mg in 5.2 grams.

To do this, we need to know the molar mass of Mg, which is approximately 24.31 g/mol.

Number of moles of Mg = mass of Mg / molar mass of Mg
Number of moles of Mg = 5.2 g / 24.31 g/mol ≈ 0.214 moles

Since the stoichiometry of the reaction indicates that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2 gas, the number of moles of H2 gas produced will be half the number of moles of Mg.

Number of moles of H2 gas = 0.214 moles / 2 ≈ 0.107 moles

Therefore, approximately 0.107 moles of H2 gas were produced.