When 4.00 g Fe and 8.00 g Cl2 react to form FeCl3, the Fe is the limiting reactant and the Cl2 is the excess reactant. What mass of Cl2 remains once the reaction stops? 2Fe + 3Cl2 → 2FeCl3

Answer 1

2Fe + 3Cl₂ ==> 2FeCl₃

4.00 g x 1 mol Fe/55.85 = 0.0716 moles Fe

8.00 g Cl₂ x 1 mol/71 g = 0.1127 moles Cl₂

mass of Cl₂ used up during the reaction = 0.0716 mol Fe x 3 mol Cl₂/2 mol Fe x 71 g Cl2/mol = 7.63 g used

mass of Cl₂ remaining = 8.00 g - 7.63 g = 0.37 g Cl₂ remaining