Question

How many grams of Fe can be produced when 5.80 g of Fe2 O3 reacts?

Answer 1

To determine the grams of Fe that can be produced when 5.80 g of Fe2O3 reacts, we need to use stoichiometry.

The balanced chemical equation for the reaction is:

Fe2O3 + 3CO → 2Fe + 3CO2

From the balanced equation, we can see that 1 mole of Fe2O3 reacts to produce 2 moles of Fe.

Step 1: Calculate the number of moles of Fe2O3:
Number of moles = mass / molar mass

Molar mass of Fe2O3 = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.7 g/mol

Number of moles of Fe2O3 = 5.80 g / 159.7 g/mol = 0.036 moles

Step 2: Calculate the number of moles of Fe using the stoichiometry:
From the balanced equation, 1 mol of Fe2O3 produces 2 mol of Fe.

Number of moles of Fe = 0.036 mol Fe2O3 × (2 mol Fe / 1 mol Fe2O3) = 0.072 mol Fe

Step 3: Calculate the mass of Fe:
Mass = number of moles × molar mass

Molar mass of Fe = 55.85 g/mol

Mass of Fe = 0.072 mol × 55.85 g/mol = 4.01 g

Therefore, when 5.80 g of Fe2O3 reacts, approximately 4.01 grams of Fe can be produced.