A 10.0L flask contains 14.1 grams of an unknown gas. If the pressure in the flask is 2.3atm at 65C,?

Question

Which of the following is a possible identity of the gas?

a. NH3

b.CO

c.O2

d.CO2

e.CS2

Answer 1

Use the equation PV = nRT

where:

P = pressure in atm

V= volume in L

n = number of moles = mass / MW(molecular weight)

R = universal gas constant = 0.0821 L-atm / K-mol

T = temperature in K = 65 C + 273.15 = 338.15 K

since the mass of gas is given....then all you need to find is the molecular weight...

PV = nRT

PV = (mass x RT) / MW

MW  = (mass x RT) / PV

MW = (14.1 g x 0.0821 L-atm/K-mol x 338.15 K ) / (2.3 atm x 10 L)

MW = 17.02 g/mole

from the choices it is clear that NH3 is the gas that has a molecular weight of 17.0 g/mole  

Answer 2

Well, 10.0L is 0.0100m³, 65°C is about 338K and 2.3atm is about 233kPa

plugging that into PV=nRT (using 8.314472 for R, which is built into my calculator)

(233000 x 0.01) / (R x 338) = 0.829mol

14.1 / 0.829 = 17gm/mol

looks like a. is the only option N having a atomic mas of about 14 and H about 1, 14 + 3x1 = 17

since atomic mass of C ≈ 12 and O ≈ 16 no combination of the two is going to be 17 and S at about 32 is out of the question