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)
@Cuanchi can you help me?
You use the ideal gas equation \[pV = nRT \] and then rearrange and solve for the number of moles of water. btw r = 0.08 \[\frac{ pV }{ TR } = \]
@photon336 So the equation would be n=(0.0821)*(275 K)/(1.2 atm)*(18.2 L)?
\[(\frac{ 1.2atm*18.2L }{ 0.08~L~mol^{-1}atm~K^{-1}*275~K } )= n_{moles}\]
@Photon336 So the rounded answer is 1 mole? And then what? Don't I have to multiply that by the molar mass?
so we have 1 mole of water vapor
Yeah. How do you find molar mass?
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 \[\frac{ CH4 }{ 2H_2O }*1~Mol~H_2O = mol~?~CH_4\]
Becuase there are 2 moles of water there should be 2 moles of CH4?
Well that's in the balanced chemical reaction. but remember we have 1 mole of water
and the molar ratio of CH4 to water is 1/2
so that's why I did \[\frac{ CH4 }{ 2H_2O }*1~Mol~H_2O = mol~?~CH_4\]
So it's .5?
yep so now you need to figure out the number of grams of CH4 you need
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 \[moles*(grams/mole) = grams \]
Ok, I don't have the proper periodic table, but 0.5 moles of CH4 is 8? Or did I do that wrong?
16 grams/mole
(0.5 mol CH4) * (16 grams/mole)= 8 grams? Right?
yes
So 8 is the molar mass?
16 is the molar mass in grams/mole
when you take the product of those two you end up getting the number of grams of CH4
oh. okay thanks for the help :D
no problem. just remember to close out your question when you're done.
Join our real-time social learning platform and learn together with your friends!