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:

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

65.2 grams

Explanation:

The balanced equation of the reaction is:

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