Given the following reaction

2Al + 3Cl2 - -> 2AlCl3

How many grams of AlCl3 aluminum chloride will be produced from 125 g of Al and 125 g of Cl2?

156 g of chloride are produced

Explanation:

We state the reaction:

2Al + 3Cl₂ → 2AlCl₃

We have both masses of each reactant so we can determine the moles and then, the limiting reagent. We convert the mass to moles:

125 g / 26.98 g/mol = 4.63 moles of Al

125 g / 70.9 g/mol = 1.76 moles of Cl₂

Ratio is 2:3. 2 moles of Al need 3 moles of chlorine to react

Therefore 4.63 moles of Al will react with (4.63.3) / 2 = 6.94 moles of Cl₂

We need 6.94 moles of Cl₂ and we only have 1.76; that's why the Cl₂ is the limiting reagent.

The stoichiometry is 3:2 so let's make a new rule of three:

3 moles of chlorine can produce 2 moles of chloride

Then, 1.76 moles of Cl₂ may produce (1.76. 2) / 3 = 1.17 moles of chloride.

We convert the moles to mass = 1.17 moles. 133.33 g / 1 mol = 156 g