Part A What is the mole fraction of O2 in a mixture of 15.1 g of O2, 8.17 g of N
ID: 507829 • Letter: P
Question
Part A
What is the mole fraction of O2 in a mixture of 15.1 g of O2, 8.17 g of N2, and 2.47 g of H2?
Part B
What is the mole fraction of N2 in a mixture of 15.1 g of O2, 8.17 g of N2, and 2.47 g of H2?
Part C
What is the mole fraction of H2 in a mixture of 15.1 g of O2, 8.17 g of N2, and 2.47 g of H2?
Part D
What is the partial pressure in atm of O2 of this mixture if it is held in a 15.20 L vessel at 14 C?
Part E
What is the partial pressure in atm of N2 of this mixture if it is held in a 15.20 L vessel at 14 C?
Part F
What is the partial pressure in atm of H2 of this mixture if it is held in a 15.20 L vessel at 14 C?
Explanation / Answer
To calculate the mole fraction of each component in the mixture, first convert the masses to moles.
To do that, use the molar mass of oxygen gas, nitrogen gas, and hydrogen gas, respectively.
molar mass O2 = 31.9988
15.1/ 31.9988 = 0.472 moles O2
molar mass N2 = 28.0134
8.17/ 28.0134 = .292 moles N2
molar mass H2 = 2.01588
2.47 / 2.01588 = 1.22 moles H2
The total number of moles present in the mixture will be
ntotal = 1.99 moles
To get the mole fraction of a component i of the mixture, all you have to do is divide the number of moles of that component by the total number of moles present.
Part A
the mole fraction of O2 = 0.472/1.99 = 0.237
Part B
the mole fraction of N2 = 0.292/1.99 = 0.147
Part C
the mole fraction of H2 = 1.22/1.99 = 0.613
Using the moles fractions, determine partial pressure by mole fraction (X) times total pressure (P).
total pressure exerted by the mixture can be calculated by Raoult's Law.
P = nRT/V
n=number of moles (1.99)
R = gas constant (0.0821)
T = temp in Kelvin (14 celcius is 287 K)
V= volume in liters (15.20 L)
P=[(1.99)(.0821)(287)]/15.20 = 3.08 atm
Part D
partial pressure in atm of O2: (0.237)*(3.08)=.730 atm
N2: (0.147)*(3.08)=.452 atm
H2: (0.613)*(3.08) = 1.88 atm