Question

Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?

Answer 1

Answer:

65.08 g.

Explanation:

• For the reaction, the balanced equation is:

2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,

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

• Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

n = mass/molar mass = (36.2 g)/(133.34 g/mol) = 0.2715 mol.

Using cross multiplication:

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass = (0.4072 mol)(159.808 g/mol ) = 65.08 g.