Thread: a couple chemistry questions
12-09-2003, 11:29 PM #1
a couple chemistry questions
Now i nkow that some of you must be better at chem than I am...
A 1.0L sample of unknown gas at 25 degrees C and 211mmhg is found to weigh 0.727g. the gas is He, NO, CO2, SO2??
Consider the combustions of CH4:
CH4(g) + 2O2 (gas) = CO2(g) + 2H2O(L). If the reaction takes place at STP, how many liters of CH4 is needed to produce 1 mol of H20 (water).
thanks in advance bros
also if you know the anwer that is great but I to have the answers but am not quite sure how to get there so if you could explain a little that would be great..
12-09-2003, 11:41 PM #2
Sorry the second part of the question the answer is 11.2L or B.
12-09-2003, 11:48 PM #3
211mmHg is the atmospheric pressure...
12-10-2003, 12:11 AM #4
haha it seems like my batteries always die when I need to use something...
also i think 1 mol = 1 L
12-10-2003, 12:19 AM #5
Can't you just convert pressure to atm and use ideal gas law?
PV=nRT, and then convert the moles that you got into the atomic weight of one of the elements.
12-10-2003, 12:34 AM #6Originally Posted by Jenseno9
12-10-2003, 07:26 AM #7
22.4L CH4/1 * 1 mol CH4/1 Mol CH4 * 1 mol H20/ 2 Mol H20
= 11.2 L
Although i wouldnt you would do this b/c water is a pure liquid and normally pure liquids and solids arent counted but oh well....
12-10-2003, 07:47 AM #8
For the second one you need to put everything in standard state so:
V1/V2 = T1/T2 * P2/P1 the combined law
1/ x = 298 K /273 K * 1 atm/ (211/760)
x = 3.932 L @ STP
then use pv=nrt
1 * 3.932 = (.727/x) * .0821 * 273
x = 4 g/mol
If you dont convert it to STP you get
211/760 * 1 = (.727/x) * .0821 * 298
x = 64 g/mol
I think the first one (4 g is correct tho)....
12-10-2003, 09:23 AM #9Originally Posted by AbsolutelyLethal
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