tywower
  • tywower
How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken at the same temperature and pressure? Show all of the work used to solve this problem. CH4 (g) + 2 O2 (g) yields CO2 (g) + 2 H2O (g)
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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tywower
  • tywower
@Abhisar
tywower
  • tywower
@Preetha
tywower
  • tywower
@sweetburger

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tywower
  • tywower
@dan815
tywower
  • tywower
@Kbug
tywower
  • tywower
@Abhisar
Abhisar
  • Abhisar
This one will also be solved using PV=nRT
Abhisar
  • Abhisar
We can see from the equation that for each mole of CH4, 2 moles of oxygen is produced. And also pressure and temperature is same. We can write that \(\sf P= \huge \frac{nRT}{8.9} = \frac{2nRT}{V}\), Here V is the volume of oxygen formed. Solve the equation for v.
Abhisar
  • Abhisar
Any problem?
tywower
  • tywower
not yet @Abhisar you may continue
Abhisar
  • Abhisar
That's it. V will be the volume of oxygen. Solve the equation for v.
tywower
  • tywower
im not sure how to solve the equation @Abhisar
Abhisar
  • Abhisar
\(\sf \huge \frac{nRT}{8.9} = \frac{2nRT}{V}\) Can you solve it for V now?
Abhisar
  • Abhisar
Cancel out the common terms on both sides.
tywower
  • tywower
4.5?
tywower
  • tywower
@dan815
tywower
  • tywower
@Abhisar
taramgrant0543664
  • taramgrant0543664
It's not 4.5 unless my math is really off, all you have to do is cancel out the like terms as someone has already shown above. And if you're having a problem with that just factor out nRT from each side and that should help you get your answer
aaronq
  • aaronq
Because "all measurements are taken at the same temperature and pressure" moles are proportionate to volume and therefore conversions between moles and L can be ignored - that is we don't need the ideal gas law. We set up a ratio (as we normally would) except using liters instead of moles, and plug in the variables we know: \(\sf \dfrac{L~of ~CH_4}{CH_4's ~coefficient}=\dfrac{L~of~H_2O}{H_2O's ~coefficient}\rightarrow \sf \dfrac{8.9~L}{1}=\dfrac{L~of~H_2O}{2}\) We solve this algebraically and obtain: \(\sf L~of~H_2O=\dfrac{2*8.9~L}{1}\)

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