The correct option is D O2 is the limiting reagent
4Fe+3O2→2Fe2O3
Moles of O2=4.832=0.15 mol
Finding the limiting reagent,
For Fe=given moles stoichiometric coefficient=0.154=0.0375
For O2=given moles stoichiometric coefficient=0.153=0.05
So, Fe is the limiting reagent.
4 mol of Fe react with 3 mol of O2
0.15 mol of Fe will react with, 34×0.15=0.1125 moles of O2
Amount of O2 reacted = 0.1125×32=3.6 g
Amount of O2 left = 4.8−3.6=1.2 g
4 moles of Fe produces 2 moles of Fe2O3
Molar mass of Fe2O3=56×2+3×16=160 g/mol
0.15 mol will produce 2×0.154 = 0.075 mol of Fe2O3
Amount of Fe2O3=0.075×160=12 g
O2 is the excess reagent.
Therefore, option (d) is the incorrect statement.