Magnesium and hydrochloric acid combine in the reaction described by this equation: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) In a laboratory experiment 45.0 g Mg is reacted with excess HCl. What is the theoretical yield? (Note: The molar masses are Mg = 24.3 g/mol; MgCl2 = 95.2 g/mol.)

m = 176.12 g

Explanation:

Let's write the reaction again:

Mg(s) + 2HCl(aq) > MgCl₂(aq) + H₂(g)

This reaction is a simple displacement reaction, in this case, the magnesium is displacing the atoms of hydrogens to form MgCl₂. 1 mole of Mg reacts with 2 moles of HCl. If the magnesium is the limiting reactant, we need to calculate the theorical yield of MgCl₂. The mole ratio between these two is 1:1 so, all we have to do is calculate the moles of magnesium and then the moles of MgCl₂:

moles Mg = 45 / 24.3 = 1.85 moles

As the mole ratio is 1:1 then:

moles Mg  = moles MgCl₂ = 1.85 moles

Now, we calculate the mass using the molecular mass:

m = 1.85 * 95.2

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