Question

Dalton's law states that the total pressure, Ptotal, of a mixture of gases in a container equals the sum of the pressures of each individual gas:

Ptotal=P1+P2+P3+…

The partial pressure of the first component, P1, is equal to the mole fraction of this component, X1, times the total pressure of the mixture:

P1=X1×Ptotal

The mole fraction, X, represents the concentration of the component in the gas mixture, so

X1=moles of component 1total moles in mixture

Question 1 ---- Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 3.50 atm . Calculate the partial pressure of each gas in the container.

Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.

Part 2 ----

A gaseous mixture of O2 and N2 contains 34.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 665 mmHg ?

Express you answer numerically in millimeters of mercury.

Answer 1

Question1.          PV = nRT
n = PV / RT = (3.50 atm) x (10.0 L) /((0.08205746 L atm/K mol) x (23.0 + 273.15 K)) = 1.440 mol total of 3 gases

(8.00 g CH4) / (16.04257 g CH4/mol) = 0.4987 mol CH4
(0.4987 mol) / (1.440 mol) x 3.50 atm = 1.21 atm CH4

(18.0 g C2H6) / (30.06924 g C2H6/mol) = 0.5986 mol C2H6
(0.5986 mol) / (1.440 mol) x 3.50 atm = 1.45 atm C2H6

3.50 atm - 1.21 atm CH4 - 1.45 atm C2H6 = 0.84 atm propane

part 2: Nitrogen is 34.8grams in 100g of O2 and N2 mixture.

Therefore amount of oxygen = 100g-34.8g

                                              = 65.2g

number of moles of oxygen = 65.2g/32g/mol

                                             = 2.0375 mol

number of moles of nitrogen = 34.8g/28g/mol

                                             = 1.24 mol

hence partial pressure of oxygen = mole fraction of oxygen x total pressure

                                                     = 2.0375 mol/ (2.0375 mol+1.24mol) x 665 mm of Hg

                                                     = 413.4 mm of Hg

Hence the partial pressure of O2 is 413.4 mm of Hg .