Reaction between Iron and oxygen to form iron (iii) oxide is written as;
4Fe (s) + 3O2(g) = 2Fe2O3(s)
1mole of iron contains 56 g, hence,
28.1 g of iron contains 28.1/56 = 0.502 moles
To get the moles of iron (iii) oxide formed we use the mole ratio
mole ratio of iron : iron (iii) oxide = 4 : 2
Therefore, moles of iron(iii) oxide will be given by 0.502×2/4 = 0.251 moles
hence, the answer is 0.251 moles of iron(iii) oxide
