How to solve this dimensional analysis/stoichometry problem
Figured it out, thanks to @M. Farooq's help.
I first did some cross multiplication of 250/24 times X/15 to find out that in 15 hours, there would be 156.25 L of methane. This is equal to 86.56 grams of methane. Using that, I found out there were 5.393 moles of methane, and multiplying that by 2 (because for every one methane it makes 2 waters) I found that there would be 10.79 mol of water when the reaction was done. This, times the molar mass of water, gives me 194.4 grams of water after the explosion.