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.

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