Can someone show me how to use PV=nRT for the question How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the work used to solve this problem.

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

Question

Can someone show me how to use PV=nRT for the question How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the work used to solve this problem. 

CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)

anonymous · Answer

Look at what you are given. You are given the pressure volume and temperature. Thus you can use the equation PV= nRT because you have every variable but n (the number of moles). R will be .08206 because your units of volume and pressure are L and atm.

(1.1)(12.5)= n(.08206)(301)
n= .55668 moles 

Now that you have the number of moles you can calculate the number of grams by multiplying the number of moles by the molar mass of methane. 

(.55668)(16.0425)= 8.93 grams