How many liters of methane gas (CH4) need to be combusted to produce 8.5 liters of water vapor, if all measurements are taken at the same temperature and pressure? Show all of the work used to solve this problem.

Question

jfraser · Answer

first things first, can you put together a balanced reaction for this combustion?

anonymous · Answer

I am not sure if am correct but this is how I would answer this question. 
The combustion reaction is: CH4 + O2 = H2O + CO2
First balance the reaction: CH4 + 2 O2 = 2 H2O + CO2
Atomic weight of CH4 is 16 and 2 H2O is 36
So you get 2.25g of water vapor for each 1g of CH4
I am not sure how many grams of water vapor there are per L at any given pressure/temp. If we assume water vapor is an ideal gas then there is 22.4 L per Mol at 0 deg C. 
We only need 8.5L so 8.5L / 22.4L = .38. We need .38 of a Mol of water vapor to get 8.5L. 
We know that 2 Mol of water vapor is 36g. So 36/2 * .38 = 6.8g of water vapor.
As we determined earlier 2.25g of water vapor is produced for each 1g of CH4. 
So 6.8 / 2.25 = 3.04
We need 3 grams of methane to get 8.5L of water vapor if water vapor is an ideal gas at STP and 0 deg C.

I apologise if this is incorrect.