Question

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.)

Answer 1

Answer:

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

m MgCl₂ = 176.12 g

This is the theorical yield

Answer 2

Answer:

Theoretical yield of MgCl2 is 176.3g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Mg(s) + 2HCl(aq) —> MgCl2(aq) + H2(g)

Step 2 :

Determination of the mass of Mg that reacted and the mass of MgCl2 produced from the balanced equation. This is illustrated below:

Molar Mass of Mg = 24.3 g/mol

Molar Mass of MgCl2 = 95.2 g/mol

From the balanced equation,

24.3 g of Mg reacted.

95.2 g of MgCl2 is produced.

Step 3:

Determination of the theoretical yield of MgCl2.

This is illustrated below:

From the balanced equation above,

24.3 g of Mg reacted to produce 95.2 g of MgCl2.

Therefore, 45g of Mg will react to produce = (45 x 95.2)/24.3 = 176.3g of MgCl2.

Therefore, the theoretical yield of MgCl2 is 176.3g