How many grams of methane gas (CH4) need to be combusted to produce 18.2 L water vapor at 1.2 atm and 275 K? Show all of the work used to solve this problem.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g)

Question

How many grams of methane gas (CH4) need to be combusted to produce 18.2 L water vapor at 1.2 atm and 275 K? Show all of the work used to solve this problem. 

CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O(g)

abb0t · Answer

Start by using the ideal gas law to get "n" which is moles.
Remember: $PV = nRT$
that means that you'll be using the tricks you learned in basic algebra to rearrange the formula and solve for "n".
P = pressure (atm)
V = volume (usually L)
n = mole
R = gas constant (you can find this value in your textbook)
T= Temperature (in Kelvins)

Once you have moles of water, use dimensional analysis to get to methane. Use your balanced chemical equation to help you in this dimensional analysis. The route you'll be taking is
$moles~of~H_2O \rightarrow moles~of~CH_4 \rightarrow grams~of~CH_4$

[HINT: The balanced equation tells you that 2 moles of $H_2O$ are produced from 1 mole of $CH_4$]

You should be able to finish the problem from here, but if you still need more help, please let me know. But don't just simply ask for the answer without trying, try and figure it out yourself first :)