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

AlCl3 + Br2 → AlBr3 + Cl2

65.2 grams

69.8 grams

72.1 grams

76.5 grams

65.2 grams

Explanation:

The balanced equation of the reaction is:

2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂

It is clear that 2 mol of AlCl₃ react with 3 mol of Br₂.

firstly, we need to calculate the number of moles of both

for AlCl₃:

number of moles = mass / molar mass = (36.2 g) / (133.34 g/mol) = 0.271 mol.

2 mol of AlCl₃ needs → 3 mol of Br₂

0.27 mol of AlCl₃ needs →? mol of Br₂

∴ the number of mol of Br₂ needed = (0.271 * 3) / 2 = 0.405 mol

for Br₂:

mass = number of moles * molar mass = (0.407 mol) * (159.8 g/mol) = 65.1 g.

So, the right choice is 65.2 grams