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.)

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

m MgCl₂ = 176.12 g

This is the theorical yield

Eduard22sly Eduard22sly · Answer 2 · 2021-12-03T17:31:06+01:00

The theoretical yield of MgCl₂ obtained from the reaction is 176.3 g

We'll begin by calculating the mass of the Mg that reacted and the mass of MgCl₂ produced from the balanced equation.

Mg(s) + 2HCl(aq) —> MgCl₂(aq) + H₂(g)

Molar mass of Mg = 24.3 g/mol

Mass of Mg from the balanced equation = 1 × 24.3 = 24.3 g

Molar mass of MgCl₂ = 95.2 g/mol

Mass of MgCl₂ from the balanced equation = 1 × 95.2 = 95.2 g

SUMMARY:

From the balanced equation above,

24.3 g of Mg reacted to produce 95.2 g of MgCl₂

Finally, we shall determine the theoretical yield of MgCl₂.

From the balanced equation above,

24.3 g of Mg reacted to produce 95.2 g of MgCl₂

Therefore,

45 g of Mg will react to produce = (45 × 95.2)/24.3 = 176.3 g of MgCl₂

Thus, the theoretical yield of MgCl₂ is 176.3 g

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