When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:

2 Al (s) + 3 Cl2 (g) - -> 2 AlCl3 (s)

b) How many moles of AlCl3 are formed?

1.5 mole

Explanation:

Step 1:

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

2Al (s) + 3Cl2 (g) - -> 2AlCl3 (s)

Step 2:

Determination of the masses of Al and Cl2 that reacted 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

From the balanced equation,

54g of Al reacted.

213g of Cl2 reacted

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 40.5g of Al will react with = (40.5 x 213) / 54 = 159.75g of Cl2.

From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

Step 4:

Determination of the number of mole in 40.5g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 40.5g

Number of mole of Al = ?

Number of mole = Mass/Molar Mass

Number of mole of Al = 40.5/27

Number of mole of Al = 1.5 mole

Step 5:

Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

2Al (s) + 3Cl2 (g) - -> 2AlCl3 (s)

From the balanced equation above,

2 moles of Al produced 2 moles of AlCl3.

Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.

1.5 moles AlCl3 will be formed

Explanation:

Step 1: Data given

MAss of Al = 40.5 grams

Mass of Cl2 = 212.7 grams

Atomic mass Al = 26.98 g/mol

Molar mass Cl2 = 70.9 g/mol

Step 2: The balanced equation

2 Al (s) + 3 Cl2 (g) - -> 2 AlCl3 (s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al = 40.5 grams / 26.98 g/mol

Moles Al = 1.50 moles

Moles Cl2 = 212.7 grams / 70.9 g/mol

Moles Cl2 = 3.0 moles

Step 4: Calculate limiting reactant

For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

Al is the limiting reactant. It will completely be consumed (1.5 moles).

Cl2 is in excess. There will react 3/2 * 1.5 = 2.25 moles Cl2

There will remain 3.0 - 2.25 = 0.75 moles Cl2

Step 5: Calculate moles AlCl3

For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

For 1.5 moles AlC we'll have 1.5 moles AlCl3

1.5 moles AlCl3 will be formed