a couple chemistry questions

[Jenseno9]

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).

a. 22.4L
b. 11.4L
c. 44.8L
d. 0.5L

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..

[Jenseno9]

atomic weights
SO2 64.07
CO 44.01
NO 30.01
HE 4.00


Sorry the second part of the question the answer is 11.2L or B.

[Jenseno9]

211mmHg is the atmospheric pressure...

[Jenseno9]

haha it seems like my batteries always die when I need to use something...

also i think 1 mol = 1 L

[rambo]

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.

[Dude-Man]

haha it seems like my batteries always die when I need to use something...

also i think 1 mol = 1 L

1 mol at standard temperature and pressure = 22.4 L

[AbsolutelyLethal]

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....

[AbsolutelyLethal]

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)....

[cb25]

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....

****...this part i actually figured out...was proud of myself until i saw someone had already answered it...haha...but the second post is way beyond me at this point...i dumped chemistry back in undergrad...no more i say!