Question

57. For the reaction shown, calculate the theoretical yield ofthe product in grams for each of the initial quantities ofreactants.2 Al(s) + 3 Cl₂(g) 2 AlCl3(s)(a) 1.0 g Al; 1.0 g Cl₂(b) 5.5 g Al; 19.8 g Cl₂(c) 0.439 g Al; 2.29 g Cl₂

Answer 1

In this question, we have to calculate the theoretical yield of AlCl3, based on the following reaction:

2 Al + 3 Cl2 -> 2 AlCl3

In this reaction, we have the following molar ratios:

2 Al = 3 Cl2

2 Al = 2 AlCl3

3 Cl2 = 2 AlCl3

We have as the initial mass of the reactants:

1.0 grams of Al and 1.0 grams of Cl2

We need to find the limiting and excess reactants before the theoretical yield, and in order to find the limiting and excess reactants, we need to calculate the number of moles we have in 1.0 grams of each

Starting with Al, the molar mass of Al is = 27g/mol

27g = 1 mol of Al

1.0g = x moles of Al

27x = 1

x = 0.37 moles of Al in 1 gram of Al

According to the molar ratio between Al and Cl2, 2 moles of Al = 3 moles of Cl2, what if we have 0.37 moles of Al:

2 Al = 3 Cl2

0.37 Al = x Cl2

2x = 1.11

x = 1.11/2

x = 0.55 moles of Cl2 will be needed to react with 0.37 moles of Al

The molar mass of Cl2 is 70.9g/mol:

70.9g = 1 mol of Cl2

1.0g = x moles

70.9x = 1

x = 1/70.9

x = 0.014 moles

We have 0.014 moles of Cl2 in this reaction, and we needed 0.55 moles of Cl2, which means that Cl2 is the limiting reactant and Al is in excess

Using the number of moles of the limiting reactant, 0.014 moles, we will find the final mass of the product

First we need to know how many moles of AlCl3 we have, based on the molar mass between Cl2 and AlCl3, 3:2

3 Cl2 = 2 AlCl3

0.014 Cl2 = x AlCl3

3x = 0.028

x = 0.028/3

x = 0.009 moles of AlCl3

Now we have the number of moles of AlCl3, the molar mass is = 133.34g/mol

133.34g = 1 mol

x grams = 0.009 moles of AlCl3

x = 0.009 * 133.34

x = 1.20 grams of AlCl3

The theoretical yield of AlCl3 for 1.0 gram of Al and 1.0 gram of Cl2 will be 1.20 grams