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) c) How many moles of the excess reactant remain at the end of the reaction?

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al 212.7 g of Cl₂

                     40.5 g of Al  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl 1 mol

                       53.057 g of Cl  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

There will remain 0.75 moles of Cl2

Explanation:

Step 1: Data given

Mass of Al = 40.5 grams

Mass of Cl2 = 212.7 grams

Molar mass Al = 26.99 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 / 26.99 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 (1.50 moles)

Cl2 is in excess. There will react 3/2 * 1.50 = 2.25 moles

There will remain 3.0 - 2.25 = 0.75 moles

There will remain 0.75 moles of Cl2

your answer would be “dry”