If 29.0 l of methane, ch4, undergoes complete combustion at 0.961 atm and 140c, how many liters of each product would be present at the same temperature and pressure

Combustion of methane is as follows;

CH₄ + 2O₂ - - > CO₂ + 2H₂O

at constant temperature and pressure, volume of gas is directly proportional to number of moles of gas. This is Avagadro's law.

since its the same temperature and pressure, molar ratio is equivalent to volume ratio.

for instance molar ratio of CH₄ to CO₂ is 1:1

since number of moles is directly proportional to volume of gas, then volume ratio of CH₄ to CO₂ is 1:1

volume of methane is 29.0 L

therefore volume of CO₂ formed is 29.0 L

volume ratio of CH₄ to H₂O is 1:2

then volume of H₂O formed is 29.0 x 2 = 58.0 L

therefore volume of H₂O formed is 58.0 L