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 26.0 g of aluminum with 31.0 g of chlorine?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer – We are given, mass of Al = 26.0 g , mass of Cl₂ = 31.0 g

Reaction - 2Al(s) + 3Cl₂(g) -----> 2AlCl₃(s)

Calculation of moles of Al and Cl₂-

We know,

Moles = given mass / molar mass

Moles of Al = 26.0 g / 26.982 g.mol^-1

                     = 0.964 moles

Moles of Cl₂ = 31.0 g / 70.906 g.mol^-1

                     = 0.437 moles

Calculation of limiting reactant –

Moles of AlCl₃ from Al

From the balanced reaction –

2 moles of Al = 2 moles of AlCl₃

So, 0.964 moles of Al = ?

= 0.964 moles of AlCl₃

Moles of AlCl₃ from Cl₂

From the balanced reaction –

3 moles of Cl₂= 2 moles of AlCl₃

So, 0.437 moles of Cl₂ = ?

= 0.291 moles of AlCl₃

So, moles of AlCl₃ is lowest from the Cl₂, so Cl₂ is limiting reactant and

Moles of AlCl₃ = 0.291 moles

Calculation of mass of AlCl₃

We know,

Mass of AlCl₃ = moles of AlCl₃ * molar mass of AlCl₃

                       = 0.291 moles of AlCl₃ * 133.34 g/mol

                       = 38.9 g

The maximum mass of aluminum chloride that can be formed when reacting 26.0 g of aluminum with 31.0 g of chlorine is 38.9 g