A mixture containing 3.01 g each of CH4(g), C2H4(g) and C4H10(g) is contained in
ID: 964475 • Letter: A
Question
A mixture containing 3.01 g each of CH4(g), C2H4(g) and C4H10(g) is contained in a 1.50 L flask at a temperature of 45 degree C. (a) Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture, atm A mixture of gases contains 7.83 g of N2, 4.73 g of H2, and 7.13 g of NH3. If the total pressure of the mixture is 5.18 atm, what is the partial pressure of each component? Consider the arrangement of bulbs shown in the drawing. Each of the bulbs contains a volume of gas as shown. Pfi2 = 675 torr, PNe = 206 torr, and PH2 = 426 torr. What is the pressure of the system when all the stopcocks are opened, assuming that the temperature remains constant? (We can neglect the volume of the capillary tubing connecting the bulbs.)Explanation / Answer
36. calculate moles of each gas,
moles of CH4 = 3.01 g/16 g/mol = 0.19 mols
moles of C2H4 = 3.01 g/28.05 g/mol = 0.11 mols
moles of C4H10 = 3.01 g/58.12 g/mol = 0.052 mols
Total moles of gas = 0.352 mols
moles fraction of,
CH4 = 0.19/0.352 = 0.54
C2H4 = 0.11/0.352 = 0.31
C4H10 = 0.052/0.352 = 0.15
Total pressure P = nRT/V = 0.352 x 0.08205 x (273+45)/1.5 = 6.123 atm
Partial pressure of,
CH4 = 0.54 x 6.123 = 3.30 atm
C2H4 = 0.31 x 6.123 = 1.90 atm
C4H10 = 0.15 x 6.123 = 0.92 atm
37. Calculate moles of each gas
moles of N2 = 7.83 g/28 g/mol = 0.28 mols
moles of H2 = 4.73 g/2 g/mol = 2.36 mols
moles of NH3 = 7.13 g/17 g/mol = 0.42 mols
Total moles = 3.06
moles fraction of,
N2 = 0.28/3.06 = 0.09
H2 = 2.36/3.06 = 0.77
NH3 = 0.42/3.06 = 0.14
So partial pressure of,
P(N2) = 0.09 x 5.18 = 0.47 atm
P(H2) = 0.77 x 5.18 = 3.99 atm
P(NH3) = 0.14 x 5.18 = 0.72 atm
38. Total pressure of the system = sum of all partial pressures
= 675 + 206 + 426
= 1307 torr
39. Rate of effusion of unknown gas/rate of effusion of O2 = sq.rt.(32 x 10^-3/molar mass of unknown gas)
62/32 = sq.rt.(32/molar mass of unknown gas)
molar mass of unknown gas = 8.52 g/mol
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