How much chlorine should be produced if 84.2 grams of aluminum chloride and 68.4 grams bromine are combined? 2alcl3 + 3br2 —> 3albr3 + 2cl2

20.27 g.

Explanation:

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.

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.

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