Part A Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unkno
ID: 1063925 • Letter: P
Question
Part A
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 4.50 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 B
A gaseous mixture of O2 and N2 contains 37.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 505 mmHg ?
Express you answer numerically in millimeters of mercury.
Explanation / Answer
Part A
Number of moles of CH4 = weight/molecular weight = 8.00/ 16 = 0.5
moles C2H6 = 18 / 30 =0.6
Ideal gas equation, n = PV/RT = 4.50 atm x 10.0 L/ 0.08206 L atm K-1 mol-1 x 296 K= 1.85 mol
moles propane = 1.85 - (0.5+0.6)= 0.75
partial pressure of CH4 = molefraction of CH4 x total pressure = 0.5/1.85 x 4.50 = 1.22 atm
partial pressure C2H6 = 0.6/1.85 x 4.50 = 1.46 atm
partial pressure C3H8 = 0.75/1.85 x 4.50 = 1.82 atm
part B
assume 100 g mixture.
oxygen = 37.8 % = 37.8 g and nitrogen = 100-37.8 = 62.2% = 62.2 g
mw O2 = 32 g/mole; moles O2 = 37.8 /32 = 1.18 moles
mw N2 = 28 g/mole; moles N2 = 62.2 /28 = 2.22 moles
mole fraction O2 = 1.18 / (1.18+2.22) = 0.35
so partial pressure O2 = mole fraction of O2 x total pressure
= 0.35 x 505 mmHg
= 175 mmHg