Question

2.5.4 moles of aluminum reacts with 80 moles of chlorine to produce aluminum chloride.How many grams of aluminum chloride can be produced?What is the EXCESS REACTANT?

Answer 1

First of all, we must write our reaction:

2Al + 3Cl2 => 2AlCl3 (don't forget to balance it)

The limiting reactant:

We work with stoichiometry,

2 x 1 mole Al --------- 3 x 1 mol Cl2

5.4 moles Al ------- x

x = 8.1 moles

According to this result, for 5.4 moles of Al, we need 8.1 moles of Cl2, but we have 80 moles, so CL2 is the excess reactant.

The limiting reactant = Al

The excess reactant = Cl2

Grams of AlCl3 produced procedure:

The molar mass of AlCl3 = 133 g/mol

2 x 1 mole Al --------- 2 x 133 g AlCl3

5.4 moles Al ---------- y = 718.2 g

Answer:

The excess reactant = Cl2

Grams of AlCl3 = 718.2 g