How many grams of CO are needed with an excess of Fe2O3 to produce 27.9 g Fe? Please show work. Fe2O3(s)+3CO(g)-->2Fe(s)+3CO2(g)

Firstly, It is considered as a stichiometry problem.
From the balanced equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂.
It is clear that the stichiometry shows that 1.0 moles of Fe₂O3 reacts with 3.0 moles of CO to give 2.0 moles of Fe and 3.0 moles of CO₂.
We must convert the grams of Fe (27.9 g) to moles (n = mass/Atomic mass).
n = (27.9 g) / (55.845 g/mole) = 0.499 ≅ 0.50 moles.
Now, we can get the number of moles of CO that is needed to produce 0.50 moles of Fe.
Using cross multiplication:
3.0 moles of CO → 2.0 moles of Fe, from the stichiometry of the balanced equation.
??? moles of CO → 0.50 moles of Fe

The number of moles of CO that will produce 0.50 moles of Fe (27.9 g) is (3.0 x 0.50 / 2.0) = 0.75 moles.
Finally, we should convert the number of moles of CO into grams (n = mass/molar mass).
Molar mass of CO = 28.01 g/mole.
mass = n x molar mass = (0.75 x 28.01) = 21.0 g.