Aluminium reacts with chloride to produce aluminium chloride if you begin with 3.2g of aluminium and 5.4g of chloride which is the limiting reactant and how many grams of aluminium chloride can be produced from the amount of limiting reactant available

A. Chlorine (Cl2) is the limiting reactant.

B. 6.77g of AlCl3

Explanation:

The balanced equation for the reaction. This is given below:

2Al + 3Cl2 - > 2AlCl3

Next, we shall determine the masses of Al and Cl2 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 Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 from the balanced equation = 3 x 71 = 213g

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

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

Thus, from the balanced equation above,

54g of Al reacted with 213g of Cl2 to produced 267g of AlCl3.

A. Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 3.2g of Al will react will react with = (3.2 x 213) / 54 = 12.52g of Cl2.

We can see that it will take a higher mass of Cl2 (i. e 12.52g) than what was given (i. e 5.4g) to react with 3.2g of Al.

Therefore, Cl2 is the limiting reactant.

B. Determination of the mass of AlCl3 produced from the amount of the limiting reactant available.

This is illustrated below:

The limiting reactant is Cl2 with a mass of 5.4g

From the balanced equation above, 213g of Cl2 reacted to produced 267g of AlCl3.

Therefore, 5.4g of Cl2 will react to produce = (5.4 x 267) / 213 = 6.77g of AlCl3.

Therefore, 6.77g of AlCl3 is produced from the amount of the limiting reactant available.