An 8.00L tank at 8.97°C is filled with 3.26g of carbon monoxide gas and 9.49g of
ID: 636725 • Letter: A
Question
An 8.00L tank at 8.97°C is filled with 3.26g of carbon monoxide gas and 9.49g of carbon dioxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits.
A 8.00 L tank at 8.97 °C is filled with 3.26 g of carbon monoxide gas and 9.49 g of carbon dioxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits. mole fraction x1 carbon monoxide partal pressure:atm mole fraction: carbon dioxide partial pressure: atm Total pressure in tank atmExplanation / Answer
Molar mass of CO,
MM = 1*MM(C) + 1*MM(O)
= 1*12.01 + 1*16.0
= 28.01 g/mol
Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol
n(CO) = mass of CO/molar mass of CO
= 3.26/28.01
= 0.1164
n(CO2) = mass of CO2/molar mass of CO2
= 9.49/44.01
= 0.2156
n(CO),n1 = 0.1164 mol
n(CO2),n2 = 0.2156 mol
Total number of mol = n1+n2
= 0.1164 + 0.2156
= 0.332 mol
a)
mole fraction of CO = mol of CO /total mol
= 0.1164 / 0.332
= 0.351
b)
mole fraction of CO2 = 1 - mole fraction of CO
= 1 - 0.351
= 0.649
c) calculate total pressure
Given:
V = 8.0 L
n = 0.332 mol
T = 8.97 oC
= (8.97+273) K
= 281.97 K
use:
P * V = n*R*T
P * 8 L = 0.332 mol* 0.08206 atm.L/mol.K * 281.97 K
P = 0.960 atm
Answer: 0.960 atm
d)
partial pressure of CO = mole fraction of CO * total pressure
= 0.351 * 0.960 atm
= 0.337 atm
e)
partial pressure of CO2 = mole fraction of CO2 * total pressure
= 0.649 * 0.960 atm
= 0.623 atm
Answer:
0.351
0.337 atm
0.649
0.623 atm
0.960 atm
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