Question

43. For the reaction shown, calculate the theoretical yield of product (in grams) for each initial amount of reactants. 

MISSED THIS? Read Section 4.4; Watch KCV 4.4, IWE 4.6 

2 Al(s) + 3 CI₂(g) → 2 AlCl₃(s) 

a. 2.0 g Al, 2.0 g Cl₂ 

b. 7.5 g Al, 24.8 g Cl₂ 

c. 0.235 g Al, 1.15 g Cl₂

Answer 1

Balanced equation:
2 Al + 3 Cl₂ ====> 2 AlCl₃
Reaction type: synthesis

Question a

mass of aluminium = 2 gm

Moles of aluminium = 2 gm / 26.98 g/mol = 0.07412 Moles

mass of Chlorine = 2 gm

Moles of Chlorine = 2 gm / 70.906 g/mol = 0.0282 Moles

Limiting reagent is Chlorine  

Moles of AlCl3 produced = 0.018804 Moles

Mass of AlCl3 produced =   0.018804 Moles x 133.34 gm/mol = 2.507 gm

Theoretical yield of the reaction is 2.507 gm

Question b

mass of aluminium = 7.5 gm

Moles of aluminium = 7.5 gm / 26.98 g/mol = 0.27796 Moles

mass of Chlorine = 24.8 gm

Moles of Chlorine = 24.8 gm / 70.906 g/mol = 0.34975 Moles

Limiting reagent is Chlorine  

Moles of AlCl3 produced = 0.23317 Moles

Mass of AlCl3 produced =  0.23317 Moles x 133.34 gm/mol = 31.09 gm

Theoretical yield of the reaction is 31.09 gm

Question c

mass of aluminium = 0.235 gm

Moles of aluminium = 0.235 gm / 26.98 g/mol = 0.008709 Moles

mass of Chlorine = 1.15 gm

Moles of Chlorine = 1.15 gm / 70.906 g/mol = 0.01621 Moles

Limiting reagent is Aluminium

Moles of AlCl3 produced = 0.008709 Moles

Mass of AlCl3 produced =  0.008709 Moles x 133.34 gm/mol = 1.161 gm

Theoretical yield of the reaction is 1.161 gm

Answer 2

Answer 3

Answer 4