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) yields CO2 (g) + 2 H2O (g)

Answer 1

The idea to answer this is use the ideal gas equation to work out how many moles of H2O you have under these conditions. In order to do that we proceed like this:

PV = nRT
P = 1.2 atm
V = 18.2 L
n = ? mol
R = gas constant = 0.0821 L atm mol^-1 K^-1
T = temp, must be in Kelvin, = 275 K
n = PV / RT
= 1.2 atm x 18.2 L / (0.0821 Latmmol^-1 K^-1 x 275 K)
= 0.9673 mol H2O
The balanced equation tells you that
2 moles H2O are produced from 1 mole CH4
Therefore moles CH4 required = 1/2 x moles H2O
= 1/2 x 0.9673 mol
= 0.4837 mol
mass CH4 = molar mass x moles
= 16.042 g/mol x 0.4837 mol
= 7.759 g
= 7.8 g (2 sig figs)
I hope this can help you greatly