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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)

Respuesta :

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

176 L liters of water vapor can be produced.

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 = [tex]\frac{mRT}{M}[/tex]

Given ,

T = 312 K

P = 0.98 atm

m = 108 g

M (CH4) = 16.04 g /mol

R = 0.0821 Latm/molK

V = [tex]\frac{mRT}{PM}[/tex]

Plugging the numbers in the  Ideal gas law we get,

V = [tex]\frac{(108 g)(0.0821 Latm/molK)(312K)}{(16.04g/mol)(0.98atm)}[/tex]

V = 175.99 L

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