Answer:
28.9 g
Explanation:
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Gather all the information in one place with molar masses above the formulas and masses below them.
[tex]M_{r}[/tex]: 159.69 28.01
Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂
Mass/g: 55.0
1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.
[tex]\text{Moles of Fe$_{2}$O$_{3}$} =\text{55.0 g Fe$_{2}$O$_{3}$} \times \frac{\text{1 mol Fe$_{2}$O$_{3}$}}{\text{159.69 g Fe$_{2}$O$_{3}$}}= \text{0.3444 mol Fe$_{2}$O$_{3}$}[/tex]
2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.
[tex]\text{Moles of CO} = \text{0.3444 mol Fe$_{2}$O$_{3}$} \times \frac{\text{3 mol CO}}{\text{1 mol Fe$_{2}$O$_{3}$}}= \text{1.033 mol CO}[/tex]
3.Use the molar mass of CO to calculate the mass of CO.
[tex]\text{Mass of CO} = \text{1.033 mol CO} \times \frac{\text{28.01 g CO} }{\text{1 mol CO}}= \textbf{28.9 g CO}[/tex]
