Fe + O2  Fe2O3 When iron rusts, it undergoes the reaction above. Determine the mass of rust, iron (III) oxide, when 9.75 g of iron and 8.60 g of O2 are combined.

Question

Make sure equation is balanced

Answer 1

4Fe + 3O₂ → 2Fe₂O₃ Balance equation

Calculate molar masses:

O₂ = 32g/mol

Fe = 55.85g/mol

Fe₂O₃ = 160g/mol

9.75g Fe x mol Fe/ 55.85 mol Fe x 2 mol Fe₂O₃ / 4 mol Fe

x 160g Fe₂O₃ / .0873 mol Fe₂O₃ = 14.0g Fe₂O₃ Calculate grams of Fe₂O₃

Do same calculation with oxygen to find grams of iron oxide and you will find iron generates smaller amount of iron oxide and that's the value you use.