If you had excess aluminum, how many moles of aluminum chloride could be produced from 15.0 gg of chlorine gas, Cl2Cl2? Express your answer to three significant figures and include the appropriate units.

Answer: 0.141mole of AlCl3 will be produced

Explanation:

2Al + 3Cl2 - > 2AlCl3

Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

Mass conc. Of Cl2 = 15g

n = Mass conc. / molar Mass = 15/71 = 0.211 mol

From the equation,

3 moles of Cl2 produced 2moles of AlCl3.

0.211mole of Cl2 will produce = (0.211x2) / 3 = 0.141mole of AlCl3

We will produce 0.141 moles of AlCl3

Explanation:

Step 1: Data given

Mass of Chlorine gas = 15.0 grams

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Al + 3Cl2 → 2AlCl3

Step 3: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 15.0 grams / 70.9 g/mol

Moles Cl2 = 0.212 moles

Step 4: Calculate moles AlCl3

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

For 0.212 moles Cl2 we'll have 2/3 * 0.212 = 0.141 moles AlCl3

We will produce 0.141 moles of AlCl3