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      

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.