Question

Based on the equation, how many grams of Br2 are required to react completely with 42.3 grams of AlCl3? AlCl3 + Br2 → AlBr3 + Cl2 66.5 grams 71.2 grams 76.1 grams 80.2 grams

Answer 1

Answer:

76.1 grams.

Explanation:

• From the balanced reaction:

2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,

2.0 moles of AlCl₃ reacts with 3.0 moles of Br₂ to produce 2.0 moles of AlBr₃ and 3.0 moles of Cl₂.

• We need to calculate the no. of moles of (42.3 g) of AlCl₃:

no. of moles of AlCl₃ = mass/molar mass = (42.3 g)/(133.34 g/mol) = 0.3172 mol.

Using cross multiplication:

2.0 moles of AlCl₃ react completely with → 3.0 moles of Br₂.

0.3172 mol of AlCl₃ react completely with → ??? moles of Br₂.

∴ The no. of moles of Br₂ = (0.3172 mol)(3.0 mol)/(2.0 mol) = 0.4759 mol.

∴ The amount of grams of Br₂ are required to react completely with 42.3 grams of AlCl₃ = (no. of moles of Br₂)(molar mass) = (0.4759 mol)(159.81 g/mol) = 76.05 g ≅ 76.1 g.