When 60.0g of aluminum reacts with hydrochloric acid, how many grams of aluminum chloride are prepared? Balance the reaction first and then solve the problem

296.67 g of AlCl₃.

Explanation:

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

2Al + 6HCl —>2AlCl₃ + 3H₂

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 60 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, 60 g of Al will react to produce = (60 × 267)/54 = 296.67 g of AlCl₃

Thus, 296.67 g of AlCl₃ were obtained from the reaction.