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Chemistry 13 Online
OpenStudy (anonymous):

Chemistry Help?!?! How many grams of methane gas (CH4) need to be combusted to produce 18.2 L water vapor at 1.2 atm and 275 K? Show all of the work used to solve this problem. CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

OpenStudy (anonymous):

@Cuanchi can you help me?

OpenStudy (photon336):

You use the ideal gas equation \[pV = nRT \] and then rearrange and solve for the number of moles of water. btw r = 0.08 \[\frac{ pV }{ TR } = \]

OpenStudy (anonymous):

@photon336 So the equation would be n=(0.0821)*(275 K)/(1.2 atm)*(18.2 L)?

OpenStudy (photon336):

\[(\frac{ 1.2atm*18.2L }{ 0.08~L~mol^{-1}atm~K^{-1}*275~K } )= n_{moles}\]

OpenStudy (anonymous):

@Photon336 So the rounded answer is 1 mole? And then what? Don't I have to multiply that by the molar mass?

OpenStudy (photon336):

so we have 1 mole of water vapor

OpenStudy (anonymous):

Yeah. How do you find molar mass?

OpenStudy (photon336):

now we need to go back to the balanced equation: now we multiply the number of moles of water by the molar ratio of the two to figure out how many moles of CH4 \[\frac{ CH4 }{ 2H_2O }*1~Mol~H_2O = mol~?~CH_4\]

OpenStudy (anonymous):

Becuase there are 2 moles of water there should be 2 moles of CH4?

OpenStudy (photon336):

Well that's in the balanced chemical reaction. but remember we have 1 mole of water

OpenStudy (photon336):

and the molar ratio of CH4 to water is 1/2

OpenStudy (photon336):

so that's why I did \[\frac{ CH4 }{ 2H_2O }*1~Mol~H_2O = mol~?~CH_4\]

OpenStudy (anonymous):

So it's .5?

OpenStudy (photon336):

yep so now you need to figure out the number of grams of CH4 you need

OpenStudy (photon336):

you know you have 0.5 moles of CH4 finding the number of grams is easy: 1. Moles of CH4*(molar mass of CH4) = grams \[moles*(grams/mole) = grams \]

OpenStudy (anonymous):

Ok, I don't have the proper periodic table, but 0.5 moles of CH4 is 8? Or did I do that wrong?

OpenStudy (photon336):

16 grams/mole

OpenStudy (anonymous):

(0.5 mol CH4) * (16 grams/mole)= 8 grams? Right?

OpenStudy (photon336):

yes

OpenStudy (anonymous):

So 8 is the molar mass?

OpenStudy (photon336):

16 is the molar mass in grams/mole

OpenStudy (photon336):

when you take the product of those two you end up getting the number of grams of CH4

OpenStudy (anonymous):

oh. okay thanks for the help :D

OpenStudy (photon336):

no problem. just remember to close out your question when you're done.

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