Final answer:
To find the moles of Fe₂O₃ produced from 18.0 g of Fe, convert the mass of Fe to moles and apply the stoichiometric ratio from the chemical reaction. The result is 0.161 moles of Fe₂O₃.
Step-by-step explanation:
To calculate how many moles of Fe₂O₃ will be produced from 18.0 g of Fe, given that O₂ is in excess, we utilize the molar mass of Fe and the stoichiometric relationship from the balanced chemical reaction between iron and oxygen, which forms iron (III) oxide.
First, we convert the mass of Fe to moles by dividing by its molar mass (55.85 g/mol): 18.0 g Fe ÷ 55.85 g/mol = 0.322 moles of Fe. Next, we use the balanced equation for the formation of Fe₂O₃, which shows that 4 moles of Fe produce 2 moles of Fe₂O₃. This gives us a molar ratio of 2:1 for Fe to Fe₂O₃. So, we divide the moles of Fe by 2 to find the moles of Fe₂O₃: 0.322 moles Fe ÷ 2 = 0.161 moles of Fe₂O₃. This is the amount of Fe₂O₃ produced from 18.0 g of Fe when oxygen is in excess.