Question

2Al(s)+3Cl2​(g)→2AlCl3​(s)

Given the above reaction, if you have 10.0 grams of Aluminum and Chlorine gas, what is the theoretical yield of AlCl₃ for the reaction?

Answer 1

Balanced chemical reaction is

2Al_(s) + 3Cl_2(g)    → 2AlCl_3(s)

Gm of Al = 10 gm

Molar mass of Al = 26.981539 g/mol

Gm of Cl₂ = 10 gm

Molar mass of Cl₂ = 70.906 g/mol

No. of moles = gm of compound / molar mass

Moles of Al = 10 g / 26.961539 g/mol = 0.3706 moles

Moles of HCl = 10 / 70.906 = 0.141 moles

According to balanced chemical reaction 2 mole of Al react with 3 mole of Cl₂ molar ratio between Al to Cl₂ is 2:3 therefore to react with 0.3706 mole of Al required Cl₂ = 0.3706 X 3 / 2 = 0.5559 moles Cl₂ but Cl₂ given only 0.141 moles therefore Cl₂ is limiting reactant.

Limiting reactant = Cl₂

According to balanced chemical reaction 3 mole of Cl₂ produce 2 mole of AlCl₃ molar ratio between Cl₂ to AlCl₃ is 3:2 therefore 0.141 mole of Cl₂ produce AlCl₃ = 0.141 X 2 / 3 = 0.094 moles

Mole of AlCl₃ produced = 0.094 mole

Molar mass of AlCl₃ = 133.34 g/mol

Gm of compound = no. of moles X molar mass

Gm of AlCl₃ formed = 0.094 X 133.34 = 12.53 gm

Gm of AlCl₃ formed = 12.53 gm

Theoretical yield of AlCl₃ = 12.53 gm