Dalton\'s law states that the total pressure. P_total. of a mixture of gases in
ID: 966399 • Letter: D
Question
Dalton's law states that the total pressure. P_total. of a mixture of gases in a container equals the sum of the pressures of each individual gas: P_total = P_1 + P_2 P_3 + The partial pressure of the first component. P_1. is equal to the mole fraction of this component. X_1. limes the total pressure of the mixture: Three gases (8.00 g of methane. CH_4. 18 0 g of ethane. C_2H_6. and an unknown amount of propane. C_3H_8) were added to the same 10.0-L container At 23.0 degree C. the total pressure in the container is 4.70 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. A gaseous mixture of O_2 and N_2 contains 35.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 325 mmHg ? Express you answer numerically in millimeters of mercury.Explanation / Answer
PV = nRT
4.7 * 10 = n * 0.0821 * 296
n = 1.934 moles
number of moles of CH4 = 8/16 = 0.5
number of moles of C2H6 = 18/30 = 0.6
number of moles of C3H8 = 1.934 - 0.6 - 0.5 = 0.834
P total * mole fraction = Partial pressure of component
P total * X CH4 = P CH4
4.7 * 0.5/1.934 = p CH4
P CH4 = 1.215
P c2H6 = X C2H6 * P total
P C2H6 = 0.6/1.934 * 4.7 = 1.458
P C3H8 = X C3H8 * P total
P C3H8 = 0.834/1.934 * 4.7 = 2.027
Total mass of Gas mixture = 32 + 28 = 60
mass of N2 gas = 60 * 35.8/100 = 21.48 g
mass of O2 gas = 60 * (100-35.8)/100 = 38.52 g
P O2 = X O2 * P total
P O2 = (38.52/32)/((38.52/32)+(21.48/28)) * 325
P O2 = 198.498 mm Hg
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