The equation for the complete combustion CH₄ is: CH₄ + 2O₂ → CO₂ + 2H₂O If 3.70 moles of methane reacted with 5.40 moles of oxygen, what is the limiting reactant in the formation of water?

CH₄ + 2O₂ → CO₂ + 2H₂O

from equation 1 mol 2 mol

given 3.70 mol 5.40 mol

1 mol CH₄ needs 2 mol O₂

3.70 mol CH₄ need x mol O₂

x=3.70*2/1=7.40 mol O₂

So, we see that 3.70 mol CH₄ need 7.40 mol O₂, but only 5.40 mol O₂ given (it is not enough).

So, O₂ is the limiting reactant.