I need help with a couple Chemistry problems. 4 to be exact, but I will post the seperatley. They are apart of a study guide these are 4 out of 20. I have answered the other 16 but I am struggling with these. I have no idea how to do them. Please help.
A 10.0 L steel tank contains propane gas, C3H8 at 10˚C and a pressure of 1,750 mm HG. How many grams of propane are found with in the tank?
2007-05-07 15:23:35 · 2 answers · asked by shane200388 1 in Science & Mathematics ➔ Chemistry
Use the ideal gas law: pV=nRT
p = pressure (1,750 mm Hg)
V = volume (10.0 L)
n = moles (unknown)
R = gas constant (62.3637 (L · mmHg)/(K · mol))
T = temperature (10˚C or 283.15 K)
Other neccesary information:
molecular weight of propane = 44.11 g/mol
Solve for moles and then convert moles to grams of propane.
n = pV/RT = (1750*10.0)/(62.3637*283.15) = 0.991 mol
0.991 mol * 44.11 g/mol = 43.7 grams of propane
GLUCK to ya!
2007-05-07 15:43:44 · answer #1 · answered by the_rethan 2 · 0⤊ 0⤋
(10.0 L)(293/283)(1,750/760)(44 g/22.4 L) â 46.829 g
2007-05-07 15:40:18 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋