The coefficients in the balanced reaction
4 Fe + 3 O₂ → 2 Fe₂O₃
represent the ratio of moles of each reactant to each product. That is, for every 4 moles of Fe and 3 moles of O₂, 2 moles of Fe₂O₃ are produced.
First convert the given amount of Fe into moles. Fe has a molar mass of 55.845 g/mol, so you have
42.7 g × (1/55.845 mol/g) = 0.765 mol
of Fe.
Find how many moles of Fe₂O₃ are produced given this much Fe and an abundant supply of O₂. Multiply the coefficients of the balanced reaction by 0.765/4 :
0.765 Fe + 0.573 O₂ → 0.382 Fe₂O₃
Now convert the amount of Fe₂O₃ to grams. O has a molar mass of 15.999 g/mol, so Fe₂O₃ has a total molar mass of about
2 (55.845 g/mol) + 3 (15.999 g/mol) = 159.687 g/mol
Then the reaction would produce about
0.382 mol × (158.687 g/mol) ≈ 61.0 g
of Fe₂O₃.
