Question
The partial pressure of CH4(g) is 0.185 atm and that of O2(g) is 0.300 atm in a mixture of the two gases.
a) What is the mole fraction of each gas in the mixture?
b) If the mixture occupies a volume of 11.5 L at 65 degress C, calculate the total number of moles of gas in the mixture.
a) partial pressure CH4 = 0.185 atm.
partial pressure O2 = 0.300
total pressure = 0.185 + 0.300
X=mol fraction CH4 = pressure CH4/total pressure.
X=mol fraction O2 = pressure O2/total pressure.
b)Use PV = nRT
Use total P, you have V, n is what you solve for, R is the gas constant (0.08205 L*atm/mol*K) and T is 65 (don't forget to change to Kelvin). Using total pressure will get total n.
a) What is the mole fraction of each gas in the mixture?
b) If the mixture occupies a volume of 11.5 L at 65 degress C, calculate the total number of moles of gas in the mixture.
a) partial pressure CH4 = 0.185 atm.
partial pressure O2 = 0.300
total pressure = 0.185 + 0.300
X=mol fraction CH4 = pressure CH4/total pressure.
X=mol fraction O2 = pressure O2/total pressure.
b)Use PV = nRT
Use total P, you have V, n is what you solve for, R is the gas constant (0.08205 L*atm/mol*K) and T is 65 (don't forget to change to Kelvin). Using total pressure will get total n.
Answers
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