Answer:
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.
