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X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

12.0 grams of CH 4 has a pressure of 2.5 atm and a temperature of 30 C. Calculate the volume of the gas. 12.0 g CH 4 (1mole / 16.05 g) = .747 mole CH 4 (2.5atm)(x) = (.747 mole CH 4 )(.0821 L atm /mol K)(303K) X = 7.4 L CH 4.

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X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

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  1. 12.0 grams of CH4 has a pressure of 2.5 atm and a temperature of 30 C. Calculate the volume of the gas.

  2. 12.0 g CH4 (1mole / 16.05 g) = .747 mole CH4(2.5atm)(x) = (.747 mole CH4)(.0821 L atm /mol K)(303K)X = 7.4 L CH4

  3. A sample of carbon monoxide has a volume of 20.0 L at 25 C and 619 kPa. Calculate the mass of the carbon monoxide gas.

  4. X = (619kPa)(20.0L)/(8.314)(298K)X = 5.00 mol CO5.00 mol CO (28.01 g/1mol) = 140.05 g

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