Question

How many liters of water vapor can be produced if 108 grams of methane gas (CH4) are combusted at 312 K and 0.98 atm? Show all of the work used to solve this problem. CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

Answer 1

PV=mRT/M(.98)V=(108)(.0821)(312)/(16.04)(.98)V=172.47V=176.0 L CH4

Answer 2

176 L liters of water vapor can be produced.

Step-by-step explanation:

Using the Ideal gas law,

PV = nRT, where P, T, n, R and T stand for pressure, volume, moles, universal gas constant and temperature respectively.

n = m/M, where n, m and M stand for moles, mass and molar mass respectively.

PV =
`(mRT)/(M)`

Given ,

T = 312 K

P = 0.98 atm

m = 108 g

M (CH4) = 16.04 g /mol

R = 0.0821 Latm/molK

V =
`(mRT)/(PM)`

Plugging the numbers in the Ideal gas law we get,

V =
`((108 g)(0.0821 Latm/molK)(312K))/((16.04g/mol)(0.98atm))`

V = 175.99 L