Question

For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants.
2Al(s)+3Cl2(g)→2AlCl3(s)

1) 7.7 g Al, 24.4 g Cl2

Express your answer using three significant figures.

2)0.230 g Al, 1.15 g Cl2

Express your answer using three significant figures.

Answer 1

2Al(s)+3Cl2(g)→2AlCl3(s)

no of moles of Al = W/G.A.Wt = 7.7/27   = 0.285 moles

no of moles of Cl2 = W/G.A.Wt     = 24.4/71   = 0.343 moles

2 moles of Al react with 3 moles Cl2

0.285 moles of Al react with = 3*0.285/2   = 0.4275 moles of Cl2

Cl2 is limiting reactant

3 mole of Cl2 react with Al to gives 2 moles of AlCl3

0.343 moles of Cl2 react with Al to gives = 2*0.343/3   = 0.229 moles of AlCl3

mass of AlCl3 = no of moles * gram molar mass

                        = 0.229*133.34 = 30.5g >>>> the theoretical yield of product

2.

2Al(s)+3Cl2(g)→2AlCl3(s)

no of moles of Al = W/G.A.Wt = 0.23/27   = 0.0085 moles

no of moles of Cl2 = W/G.A.Wt     = 1.15/71   = 0.0162 moles

2 moles of Al react with 3 moles Cl2

3 moles of Cl2 react with 2 moles of Al

0.0162 moles of Cl2 react with = 2*0.0162/3 = 0.0108 moles of Al

Al is limiting reactant

2 moles of Al react with Cl2 to gives 2 moles of AlCl3

0.0085 moles of Al react with Cl2 to gives = 2*0.0085/2   = 0.0085 moles of AlCl3

mass of AlCl3 = no of moles * gram molar mass

                        = 0.0085*133.34   = 1.14g >>>> the theoretical yield of product