Answer:
20.27 g.
Step-by-step explanation:
- From the balanced reaction:
2AlCl₃ + 3Br₂ → 3AlBr₃ + 2Cl₂,
2.0 mole of AlCl₃ reacts with 3.0 moles of Br₂ to produce 3.0 mole of AlBr₃, and 2.0 mole of Cl₂.
- We need to calculate the no. of moles of (84.2 g) of AlCl₃ and (68.4 g) of Br₂:
no. of moles of AlCl₃ = mass/molar mass = (84.2 g)/(133.34 g/mol) = 0.6315 mol.
no. of moles of Br₂ = mass/molar mass = (68.4 g)/(159.808 g/mol) = 0.4288 mol.
AlCl₃ reacts with Br₂ with (2: 3) molar ratio,
So, 0.2859 mole (the remaining is in excess) reacts completely with 0.4288 mole of Br₂.
Using cross multiplication:
2.0 moles of AlCl₃ produce → 2.0 moles of Cl₂.
∴ 0.2859 mole of AlCl₃ produce → 0.2859 mole of Cl₂.
∴ The amount of Cl₂ produced = no. of moles x molar mass = (0.2859 mol)(70.906 g/mol) = 20.27 g.