33.0 L of methane (CH4) undergoes complete
combustion at 0.961 atm and 20◦C. How much
CO2 is formed?
Answer in units of L
How much H2O is formed?
6 answers
I shall be happy to critique you thoughts.
I thought that at fist you would put the information into PV=nRT so .961V=(1)(.08206)(20+273). the volume would then be 25.04539583 L. I am not sure though if I am even on the right track.
If you assume that the problem is asking for volume (and not grams) and that the volume of CO2 is at the same P and T, you can work it just like the H2 and O2 problem previously. That saves a lot of time.
OR, you can use PV = nRT, solve for n, and convert n to whatever T and P you want BUT there are no DIFFERENT conditions listed so this seems like a lot of extra work to me. You should get th same answer either way.
OR, you can use PV = nRT, solve for n, and convert n to whatever T and P you want BUT there are no DIFFERENT conditions listed so this seems like a lot of extra work to me. You should get th same answer either way.
so since there is a 1:1 ratio between Ch4 and co2, would the volume of CO2 formed be equal to the volume of methane?
yes.
CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(g)
If you do it the other way though, you should get the same answer (with more work).
n = PV/RT = 0.961*33.0/0.08206*293 = about 1.32 and plug that back into
V = nRT/P = 1.32*0.08206*294/0.961 = 33.0255 but it will be exactly 33.0 if I had not rounded the 1.31898 mols to 1.32.
CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(g)
If you do it the other way though, you should get the same answer (with more work).
n = PV/RT = 0.961*33.0/0.08206*293 = about 1.32 and plug that back into
V = nRT/P = 1.32*0.08206*294/0.961 = 33.0255 but it will be exactly 33.0 if I had not rounded the 1.31898 mols to 1.32.
Thank you