H2+1/2O2→H2O 2 g16 g18 g 10 gH2 required O2=80 which is not present 64 g O2 required 8 g of H2 and H2 left =2 g. Thus, O2 is the limiting reactant and H2 is excess reactant. Hence, H2O formed from 64 of O2 =1618×64=72 g=1872 mole=4 mole