Based on the equation, how many grams of Br2 are required to react completely with 29.2 grams of AlCl3?

AlCl3 + Br2 → AlBr3 + Cl2

48.7 grams

52.6 grams

56.7 grams

61.3 grams

52.6 grams.

Explanation:

From the balanced reaction:

2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,

It is clear that 2 mol of AlCl₃ react with 3 mol of Br₂ to produce 2 mol of AlBr₃ and 3 mol of Cl₂.

We need to calculate the no. of moles of 29.2 g AlCl₃:

n = mass/molar mass = (29.2 g) / (133.34 g/mol) = 0.219 mol.

Using cross multiplication:

2 mol of AlCl₃ react completely with → 3 mol of Br₂, from stichiometry.

0.219 mol of AlCl₃ react completely with →? mol of Br₂.

∴ The no. of moles of Br₂ are required to react completely with 29.2 grams (0.219 mol) of AlCl₃ = (0.219 mol) (3 mol) / (2 mol) = 0.3285 mol.

∴ The mass of Br₂ are required to react completely with 29.2 grams (0.219 mol) of AlCl₃ = (no. of moles) * (molar mass) = (0.3285 mol) * (159.808 g/mol) = 52.5 g ≅ 52.6 g.