For the reaction ch4 + 2o2 → co2 + 2h2o, how many grams of water are produced from thecombustion of 6.35 moles of methane?

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

So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7

that 's what I think