If 0.240 mol of methane (ch4) reacts completely with oxygen, what is the final yield of water in grams?

Reaction of Methane with Oxygen is as follow,

CH₄ + 2 O₂ → CO₂ + 2 H₂O

According to this equation,

1 mole of CH₄ reacts with oxygen to produce = 36 g (2 mole) of H₂O

So,

0.240 mole of CH₄ on oxidation will produce = X g of H₂O

Solving for X,

X = (36 g * 0.240 mol) : 1 mol

X = 8.64 g of H₂O

Result:

If 0.240 mol of methane (CH₄) reacts completely with oxygen, 8.64 g of H₂O is produced.