Can someone please help me calculated how many grams of methane (CH4) are in a sealed 80 mL flask at room temperature (22C) and 780 mm Hg of pressure.
2007-06-20 05:47:05 · 7 answers · asked by Tata 1 in Science & Mathematics ➔ Chemistry
assume the gas is ideal. in which case..
P x V = n x R x T
P = pressure = 780 mm Hg
V = volume = 80 ml = 0.08 L
n = # moles
R = ideal gas constant
T = Temperature = 22°C = (273.15 + 22) °K = 295.15 °K
watch units!
look up ideal gas constant with units of mmHg, L, and °K here...
http://en.wikipedia.org/wiki/Gas_constant
use R = 62.3637 L x mmHg / mole °k
solve....
n = PV/RT = (780 mmHg) x (0.08 L ) / [ (62.3637 Lx mmHg / mole °k) x (295.15 °k)]
= (780) x (.080) / [62.3637 x 295.15)] moles
= 0.00339 moles = 3.39 x 10^-3 moles
moles = mass / molecular weight
mass = moles x molecular weight
mass = 3.39 x 10^-3 moles x 16 g / mole = 0.054 grams
2007-06-20 05:58:48 · answer #1 · answered by Dr W 7 · 2⤊ 0⤋
ok, you will need to use the "Ideal Gas Law". This isn't hard, nor simple, you just need to remember the rules.
*Change "n" to moles first
*Change temp. to kelven C=K+273 * I would also change mL into L, since the ideal volume is 22.4L/mol.
K=C-273
Pv = nRT
You don't have grams of methane.---> R = PV/T R = 760*22.4/273. This is the Ideal constant, if they give you the pressure in mercury, use the standard pressure of mercury.
(780)(.08) = n(62.36)(295). Now it is an algebra problem.
Solve for n which is: 0.003 MOLES of CH4 because n stands for moles of gas. Just convert the moles to grams......
0.003mol CH4 =16gCH4/1mol CH4 = 0.048g CH4
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capabilities that can emulate in such a subject. It is not difficult, you just need a view it at a different angle. You are probably approaching it differently and it probably is easier than it seems. Keep working relentless devotion and effort and will see what blossoms.... A profound knowledge that wasn't there before, then, you will begin only to understand to break the barrier that defies your abilities. You can do it! You can do ANYTHING you set your mind to! :)
2007-06-20 13:51:34 · answer #2 · answered by Anonymous · 0⤊ 2⤋
PV=nRT
P, V, R, and T are known. Calculate n. Then multiply by the molecular weight of methane to get mass.
2007-06-20 12:56:46 · answer #3 · answered by Michael C 7 · 0⤊ 0⤋
Ideal Gas Law, then moles to grams.
Some people cannot draw, some cannot do chemistry. A talentless artist can sell its work sheathed in grandiloquent propagandistic swill (e.g., Modern Art), but a talentless chemist is a bust. Empirical reality is not labile toward majority vote.
2007-06-20 12:53:39 · answer #4 · answered by Uncle Al 5 · 0⤊ 2⤋
Use the Ideal Gas Law.
I agree with Uncle Al, anything and everything needs effort.
2007-06-20 12:58:46 · answer #5 · answered by ag_iitkgp 7 · 0⤊ 1⤋
use the ideal gas law, PV=nRT
n=PV/RT
n=(780/760)(0.08)/(0.08205)(295.3) = .003388mol CH4 *16g/mol = 0.054 grams
2007-06-20 12:56:16 · answer #6 · answered by Kyle M 2 · 0⤊ 1⤋
pv=nRt
Solve for n.
Use n (Number of moles) to determine grams.
2007-06-20 12:55:32 · answer #7 · answered by Greg 3 · 0⤊ 0⤋