Twenty five grams of iron 3 oxide react with an excess of carbon monoxide to form 15 g of fe. carbon dioxide is the other product. calculate the theoretical yield and the percent yield of iron.

Theoretical mass = 17.42 g

Percent yield of Fe = 86.11%

The equation for the reaction between iron (iii) oxide and carbon monoxide is given by;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

We are required to calculate the theoretical yield and the percentage yield of Iron.

Step 1: Moles of iron (iii) oxide

Moles are given by dividing the mass of the compound by the molar mass.

Molar mass of Iron(iii) oxide = 159.69 g/mol

Moles of Iron(III) oxide = 25 g ÷ 159.69 g/mol

                                     = 0.156 moles

Step 2: Moles of Iron produced

From the equation 1 mole of Iron(iii) oxide reacts to produce 2 moles of Fe.

Therefore, the mole ratio of Fe₂O₃ to Fe is 1 : 2.

Thus, moles of Fe = Moles of Fe₂O₃ × 2

                              = 0.156 moles × 2

                              = 0.312 moles

Step 3: Theoretical mass of iron produced

To calculate the mass of iron we multiply the number of moles of iron with the relative atomic mass.

Relative atomic mass = 55.845

Mass of iron = 0.312 moles × 55.845

                    = 17.42 g

Step 4: Percent yield of iron

% yield = (Actual mass ÷ Theoretical mass)×100

            = (15 g ÷ 17.42 g) × 100 %

            = 86.11%

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