(a) Moles of O2 = mass / molar mass = 2.43 / 32.0 = 0.0759 mol According to Dalt
ID: 699688 • Letter: #
Question
(a)
Moles of O2 = mass / molar mass = 2.43 / 32.0 = 0.0759 mol
According to Dalton's law of partial pressures,
Total pressure = Sum of partial pressures
SO,
Ptotal = PCH4 + PO2
Ptotal = ( nCH4 R T / V ) + ( nO2 R T/ V )
Ptotal = (nCH4 + nO2) * R T / V
1.41 = ( nCH4 + 0.0759 ) * 0.0821 * (273.15 + 78) / 6.40
nCH4 = 0.237
mCH4 = 0.237 * MCH4
mCH4 = 0.237 * 16.0
mCH4 = mass of methane = 3.79 g.
(b)
Moles of CO2 = 3.34 / 44.0 = 0.0759 mol
Moles of Xe = 20.1 / 131.2 = 0.153
Mole fraction of CO2 = 0.0759 / ( 0.0759 + 0.153 ) = 0.332
Mole fraction of Xe = 0.153 / (0.153 + 0.0759) = 0.668
According Dalton's law,
Partial pressure = total pressure * mole fraction
pCO2 = 717 * 0.332 = 238.0 mmHg
pXe = 717 * 0.668 = 479.0 mmHg
(c)
Total pressure of mixture = 160 + 500 = 660 mmHg
Mole fraction of O2 = partial pressure of O2 / total pressure = 160 / 660 = 0.242
Mole fracion of CH4 = 500 / 660 = 0.758
Explanation / Answer
A mixture of oxygen and methane gases is maintained in a 6.40 L flask at a pressure of 1.41 atm and a temperature of 78 °C. If the gas mixture contains 2.43 grams of oxygen, the number of grams of methane in the mixture is 8- Submit Answer Retry Entire Group 9 more group attempts remaining A mixture of carbon dioxide and xenon gases, at a total pressure of 717 mm Hg, contains 3.34 grams of carbon dioxide and 20.1 grams of xenon. What is the partial pressure of each gas in the mixture? mm Hg mm Hg Pco, Xe Submit Answer Retry Entire Group 9 more group attempts remaining A mixture of oxygen and methane gases contains oxygen at a partial pressure of 160 mm Hg and methane at a partial pressure of 500 mm Hg. What is the mole fraction of each gas in the mixture?
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