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Dalton’s Law of Partial Pressures and Gas Law Stoichiometry

Dalton’s Law of Partial Pressures and Gas Law Stoichiometry. Dalton’s Law of Partial Pressures. The pressure of each gas in a mixture is called the partial pressure of that gas.

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Dalton’s Law of Partial Pressures and Gas Law Stoichiometry

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  1. Dalton’s Law of Partial Pressuresand Gas Law Stoichiometry

  2. Dalton’s Law of Partial Pressures The pressure of each gas in a mixture is called the partial pressure of that gas. Dalton’s Law states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each of the component gases. Ptotal = P1 + P2 + P3 + . . .

  3. Ptotal = P1 + P2 + P3 + . . . PH2 = 2.4 atm PHe = 6.0 atm Ptotal = 2.4 + 6.0 = 8.4 atm

  4. If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3: 2 atm + 1 atm = 6 atm + 3 atm 4 3 2 1

  5. The partial pressure of each gas is equal to the mole fraction (X) of each gas times the total pressure Moles gas x Ptotal = Pgas Total Moles Mole fraction is like a percent

  6. Sample Problem #1 A mixture of He, Ne and Ar gases have a total pressure of 790 mmHg. If there is 15% Ar, 60% He and 25% Ne, what is the partial pressure of each gas? Ptotal = PHe + PNe + PAr = 790 mmHg 15% Ar = (0.15)(790 ) = 118.5 mmHg + 60% He = (0.60)(790) = 474 mmHg + 25% Ne = (0.25)(790) = 197.5 mmHg P total = 790 mmHg

  7. Sample Problem #2 The partial pressure of CO2 in a mixture of gases is 0.8 atm. If the total pressure is 1.05 atm, what is the mole fraction of CO2 in the mixture? PCO2 = 0.8atm Ptotal = 1.05atm (Mole Fraction )Ptotal = PCO2 XCO2 (1.05atm) = 0.8 atm XCO2 = 0.8 atm = 0.76 1.05 atm

  8. Gas Collected by Water Displacement Agas collected by water displacement is not pure, but always mixed with water vapor. Some molecules at the surface of water always evaporate and exert a pressure called a vapor pressure. • Therefore, you must subtract the water –vapor pressure from the total pressure to get the pressure just from the gas.

  9. Sample Problem #3 50mls of CH4 was collected over water at 25oC and 748mmHg. What is the volume of the gas at STP? (The vapor pressure of water is 23.8mmHg) Ptotal = PCH4 + PH2O 748mmHg = PCH4 + 23.8mmHg PCH4 = 724.2 mmHg (dry gas) P1V1 = P2V2 (50mls)(724.2mmHg) = (760mmHg)V2 T1 T2 298K 273K Solving for V2 = 43.6 mL

  10. Gas Law Stoichiometry What volume of oxygen will form at 25oC and 1.3atm when 50 grams of KClO3 decomposes according to the following equation? 2KClO3 2KCl + 3O2 First you find out how many moles of oxygen gas will form using stoichiometry = 0.61 moles O2 50 g KClO3 1 mole KClO3 3 moles O2 122.55 g KClO3 2 moles KClO3 Then you use the Ideal gas law to get the volume. You cannot use the 22.4 L per 1 mole relationship because you are not at STP!!!! (1.3atm)V = (0.61 moles) (0.0821 L.atm/K.mole) (298K) V = 11.5 Liters

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