Answer:
228.6 g of H₂O
Explanation:
the balanced equation for the combustion of methane is as follows
CH₄ + 2O₂ ---> CO₂ + 2H₂O
molar ratio of CH₄ to H₂O is 1:2
when 1 mol of CH₄ reacts with excess O₂, 2 mol of H₂O is formed
therefore when 6.35 mol of CH₄ reacts - 2 x 6.35 mol = 12.7 mol of H₂O is formed
therefore mass of H₂O formed is - 12.7 mol x 18 g/mol = 228.6 g of H₂O is formed
