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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)
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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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1This one will also be solved using PV=nRT

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1We 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.

tywower
 one year ago
Best ResponseYou've already chosen the best response.0not yet @Abhisar you may continue

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1That's it. V will be the volume of oxygen. Solve the equation for v.

tywower
 one year ago
Best ResponseYou've already chosen the best response.0im not sure how to solve the equation @Abhisar

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1\(\sf \huge \frac{nRT}{8.9} = \frac{2nRT}{V}\) Can you solve it for V now?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Cancel out the common terms on both sides.

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0It'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
 one year ago
Best ResponseYou've already chosen the best response.0Because "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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