What is the mass in grams of magnesium chloride that can be produced from 7.52 moles of chlorine gas reacting with excess magnesium? Mg(s) + Cl₂(g) → MgCI₂(s)

Answer 1

In our question, we can see that Mg(s) is in excess, so for this question, we do not need to worry about Mg (s). We start with 7.52 mols of Cl₂(g), since we are starting with mols, we do not need to find the molar mass of chlorine gas. We need to end with grams of MgCl₂ (s). Since we need to end with grams, we need the molar mass of MgCl₂ (s).

To find the molar mass of MgCl₂ (s), we need to find the molar mass of Mg and Cl from the periodic table.

Mg - 24.32 g/mol

Cl - 35.45 g/mol

MgCl₂ (s) - 24.32 g/mol + 2(35.45 g/mol) = 95.22 g/mol

Now, we start with 7.52 mol Cl₂(g).

7.52 mol Cl₂(g)

For every one mol of 7.52 mol Cl₂(g) we have one mol of MgCl₂(s). So our equation becomes

7.52 mol Cl₂ * (1 mol MgCl₂)/(1mol Cl₂)

Lastly, we know the molar mass of MgCl₂(s) is 95.22 g/mol so our full equation is

7.52 mol Cl₂ * (1 mol MgCl₂)/(1mol Cl₂) * (95.22g MgCl₂)/(1 mol MgCl₂) = 716.05 g MgCl₂