<h3>
Answer:</h3>
Theoretical mass = 17.42 g
Percent yield of Fe = 86.11%
<h3>
Explanation:</h3>
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%
= 10 molecules NH₃