Answer:
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
