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tywower

  • one year ago

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)

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  1. tywower
    • one year ago
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    @Abhisar

  2. tywower
    • one year ago
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    @Preetha

  3. tywower
    • one year ago
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    @sweetburger

  4. tywower
    • one year ago
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    @dan815

  5. tywower
    • one year ago
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  6. tywower
    • one year ago
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    @Abhisar

  7. Abhisar
    • one year ago
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    This one will also be solved using PV=nRT

  8. Abhisar
    • one year ago
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    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.

  9. Abhisar
    • one year ago
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    Any problem?

  10. tywower
    • one year ago
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    not yet @Abhisar you may continue

  11. Abhisar
    • one year ago
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    That's it. V will be the volume of oxygen. Solve the equation for v.

  12. tywower
    • one year ago
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    im not sure how to solve the equation @Abhisar

  13. Abhisar
    • one year ago
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    \(\sf \huge \frac{nRT}{8.9} = \frac{2nRT}{V}\) Can you solve it for V now?

  14. Abhisar
    • one year ago
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    Cancel out the common terms on both sides.

  15. tywower
    • one year ago
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    4.5?

  16. tywower
    • one year ago
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    @dan815

  17. tywower
    • one year ago
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    @Abhisar

  18. taramgrant0543664
    • one year ago
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    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

  19. aaronq
    • one year ago
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    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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