The combustion reaction for methane is shown below: CH4 (9) + 2O2 (g) → CO2 (g) + 2H20 (9) + energy
If 64.0 g of methane is reacted with excess oxygen gas, what is
the pressure exerted by water vapor assuming the reaction
went to completion in a 24.0 L engine at 515 K?

P =  14.1 atm    

Explanation:

Given data:

Mass of methane = 64 g

pressure exerted by water vapors = ?

Volume of engine = 24.0 L

Temperature = 515 K

Solution:

Chemical equation:

CH₄ + 2O₂      →      CO₂ + 2H₂O + energy

Number of moles of methane:

Number of moles = mass / molar mass

number of moles = 64 g/ 16 g/mol

Number of moles = 4 mol

Now we will compare the moles of water vapors and methane.

              CH₄          :            H₂O  

                 1            :             2

                 4            :         2/1×4 = 8 mol

Pressure of water vapors:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L

P = 338.25 atm.L/ /  24.0 L

P =  14.1 atm      

yes the answer is total mass in unchanged   : )