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