Assuming 54.81 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be produced if the reaction will only produce 66.93 % yield?

181.39g of AlCl3 is produced

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3CuCl2•2H2O + 2Al → 2AlCl3 + 6H2O + 3Cu

Next, we shall determine the mass of Al that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:

Molar mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Summary:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Next, we shall determine the theoretical yield of AlCl3. This can be achieved as shown below:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Therefore, 54.81g of Al will react to produce = (54.81 x 267) / 54 = 271.01g of AlCl3.

Therefore, the theoretical yield of AlCl3 is 271.01g.

Finally, we shall determine the actual yield of AlCl3 produced from the reaction.

This can be obtain as follow:

Percentage yield of AlCl3 = 66.93%

Theoretical yield of AlCl3 = 271.01g

Actual yield of AlCl3 = ?

Percentage yield = Actual yield/Theoretical yield x 100

66.93% = Actual yield / 271.01g

Actual yield = 66.93% x 271.01

Actual yield = 66.93/100 x 271.01g

Actual yield = 181.39g.

Therefore, 181.39g of AlCl3 is produced from the reaction.