Answer:
29.47 g of AlCl₃.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2Al + 3Cl₂ —> 2AlCl₃
Next, we shall determine the mass of Al that reacted and the mass of AlCl₃ produced from the balanced equation. This can be obtained as follow:
Molar mass of Al = 27 g/mol
Mass of Al from the balanced equation = 2 × 27 = 54 g
Molar mass of AlCl₃ = 27 + (35.5× 3)
= 27 + 106.5
= 133.5 g/mol
Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g
SUMMARY:
From the balanced equation above,
54 g of Al reacted to produce 267 g of AlCl₃.
Finally, we shall determine the mass of AlCl₃ produced by the reaction of 5.96 g of Al. This can be obtained as follow:
From the balanced equation above,
54 g of Al reacted to produce 267 g of AlCl₃.
Therefore, 5.96 g of Al will react to produce = (5.96 × 267)/54 = 29.47 g of AlCl₃.
Thus, 29.47 g of AlCl₃ were obtained from the reaction.
