Question

If 2.112g of Fe2O3 reacts with 0.687g of Al, how much pure Fe will be produced?

Answer 1

Answer:

1.452 g

Explanation:

We are given;

• Mass of Fe₂O₃ as 2.112 g
• Mass of Aluminium as 0.687 g

We are required to calculate the mass of Fe produced;

Step 1: Determine the number of moles of each reactant

Moles = Mass ÷ Molar mass

Molar mass of Fe₂O₃ is 159.69 g/mol

Moles of Fe₂O₃ = 2.112 g ÷ 159.69 g/mol

                          = 0.0132 moles

                          = 0.013 moles

Molar mass of Aluminium is 26.98 g/mol

Moles of Al = 0.687 g ÷ 26.98 g/mol

                   = 0.0255 moles

                  = 0.026 moles

Step 2: Determine the rate limiting reactant.

The equation for the reaction is given by;

Fe₂O₃ + 2Al → Al₂O₃ + 2Fe

From the reaction, 0.0255 moles of Al would require 0.026 moles × 2 (0.052 moles) of Fe₂O₃.

On the other hand, 0.013 moles of  Fe₂O₃ would require 0.026 moles of Aluminium

Therefore, Fe₂O₃ is the rate limiting reactant for the reaction.

Step 3: Moles of Fe produced

1 mole of Fe₂O₃ reacts to produce 2 moles of Fe

Therefore; Moles of Fe = Moles of Fe₂O₃ × 2

                                      = 0.013 moles × 2

                                      = 0.026 Moles

Step 4: Mass of Fe produced

Mass = Number of moles × Molar mass

Molar mass of Fe is 55.845 g/mol

Therefore;

Mass of Fe = 0.026 moles × 55.845 g/mol

                  = 1.452 g

Therefore, 1.452 g of Fe were produced.