Alright, so i think i know how to do this. Can anyone check to see if i did it right?
97 gallons to milliliters= 367183.8 ml
367183.8 ml*1g/ml= 367183.8 g of water
i got the change in temperature from fahrenheit to celsius to be 22.22 degree celsius
the specific heat of water = .00418 kJ/ degree celsius *g
next i use the equation Q=mC(delta T)
i get= (367183.8g)(.00418kJ/ degree celsius *g)(22.22 degree celsius) = 34103.88447 kJ
next i find the entalpy for the combustion of methane:
CH4 + 2O2 ---> CO2 + 2H2O
Delta H= 2(-286)+(-393.5)-(-75)
= -890.5kJ/mole of CH4
then to find grams i do 34103.88447 kJ * (1 mole CH4/890.5 kJ) = 38.297 moles CH4
38.297 moles CH4 * (16.042 g CH4/ 1 mole CH4) = 614.4 g CH4
Also for the last part when they ask for the volume of the natural gas when it correspomds to 31 degree celsius and 1.6 atm, do i use PV=nRT for that since i know the pressure, temp and the number od moles? Thanks