Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

What is the maximum mass of aluminum chloride that can be formed when reacting 15.0g of aluminum with 20.0 g of chlorine?

Answer 1

2Al(s)+3Cl₂(g) → 2AlCl₃(s)

Molar mass of Al is 27 g/mol

Molar mass of Cl₂ is = 2xAt.mass of Cl

                               = 2 x 35.5

                               = 71 g/mol

Molar mass of AlCl₃ = At.mass of Al + (3xAt.mass of Cl)

                              = 27 + (3x35.5)

                             = 133.5 g/mol

According to the balanced equation ,

2 moles of Al reacts with 3 moles of Cl₂

2x27= 54 g of Al reacts with 3x71=213 g of Cl₂

M g of Al reacts with 20.0 g of Cl₂

M = (20.0x54) / 213

   = 5.07 g

So 15.0 - 5.07 = 9.93 g of Al left unreacted, so Al is the excess reactant.

Since all the mass of Cl₂ completly reacted it is the limiting reactant.

From the balanced equation,

3 moles of Cl₂ produces 2 moles of AlCl₃

3x71= 213 g of Cl₂ produces 2x133.5 = 267 g of AlCl₃

20.0 g of Cl₂ produces N g of AlCl₃

N = (267x20.0) / 213

   = 25.1 g

So the maximum mass of aluminum chloride produced is 25.1 g