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A sample natural gas contains 8.24 moles of methane 0.421 moles of Ethelin o116moles of propane if the total pressure of the gases is 1.77atm what are the partial pressures of the gases?
Wiki User
∙ 11y ago
mole fraction of methane = (8.24)/(8.24+0.421+0.116) = 0.939
mole fraction of ethelin = (0.421)/(8.24+0.421+0.116) = 0.048
mole fraction of propane = (0.116)/(8.24+0.421+0.116) = 0.0132
Partial pressure = Total pressure * mole fraction
P(i) = P(t) * X(i)
P(methane) = 1.77 atm * 0.939 = 1.66 atm
P(ethelin) = 1.77 atm * 0.048 = 0.085 atm
P(propane) = 1.77 atm * 0.0132 = 0.0234 atm
1.66 + 0.085 + 0.0234 = 1.7684 --> 1.77 atm
Wiki User
∙ 11y ago
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A sample natural gas contains 8.24 moles of methane 0.421 moles of Ethelin 0.116moles of propane if the total pressure of the gases is 1.77atm what are the partial pressures of the gases?
First you need to find the mole ratio of the gases. So take the moles of the gas and divide it by the total moles. After you've done that, you can set that equal to the partial pressure equation, P1 divided by Ptotal. Multiple Ptotal to both sides and you will get the answer
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the presence of Ethelin gas in the air has the effect of ripening many fruits.
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