Answer to Question #118944 in General Chemistry for Abby

Question

For the reaction
?Al+?Cl2 →?AlCl3 ,
what is the maximum amount of AlCl3 which could be formed from 6 mol of Al and 7.88 mol of Cl2?
Answer in units of g.

Answer 1

The equation is

2Al + 3Cl₂ = 2AlCl₃

Chlorine is the limiting reactant, that's why the calculations of the formed product will be performed using the quantity of Cl₂

n(Cl₂) = 7.88 mol

n(AlCl₃) = 2/3 n(Cl₂) = 5.25 mol

m(AlCl₃) = n(AlCl₃)*M(AlCl₃) = 700.04 g

The maximum amount of aluminum chloride is 700.04 g