partial pressure?

Question

A mixture of 1.02 g of H2 and 2.21 g of He exerts a pressure of 0.480 atm. What is the partial pressure of each gas present in the mixture?
H2
atm
He
atm

Answer 1

M(H2)=1g/mol
M(He)=2g/mol

n(H2)=1.02g/1g/mol=1.02mol
n(He)=2.21g/2g/mol=1.105mol
ntotal=2.125mol
x(H2)=1.02/2.125=0.48
x(He)1.05/2.125=0.52

p(H2)=0.48atm*0.48=0.2304atm
p(He)=0.48atm*0.52=0.2496atm
--------------------------------------------------
P=p(H2)+p(He)=0.48atm q.e.d

Answer 2

mol(H2) = 1.02 / 2 = 0.51
mol(He) = 2.21 / 4 = 0.5525

Ratio H2 : He :: 0.51 : 0.5525
Ratio H2 : He :: 1 : 1.0833333.......
Ratio H2 : He :: 1 : 1 (Approx.)
Therefore at a pressure of 0.480 atm each gas exerts half of the pressure.
p(H2) = 0.48/ 2 = 0.24 atm
p(He) = 0.48 /2 = 0.24 atm

Answer 3

To find partial pressure, take the molar ratio and multiply it by total pressure.

Molar ratio is the number of moles of a substance divided by the total number of moles.

So for H2, number of moles is:

n = m/Mr
n = 1.02/2
n = 0.51

For He,
n = 0.5525

Total number of moles = 0.51 + 0.5525 = 1.06
So molar ratio of H2 is 0.51/1.06 = 0.48

Therefore, partial pressure of H2 is 0.48 * 0.48atm = 0.23atm

You can do the same for helium. They may as well be equal - it depends how you round it.

Answer 4

:))

H2: 1.02g -> .5 mol gas
He: 2.21g -> .5 mol gas

-> Partial pressure is equal -> P(He) = P(H) = .240 atm