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Gas Laws: Gas Stoichiometry At the conclusion of our time together, you should be able to:

Gas Laws: Gas Stoichiometry At the conclusion of our time together, you should be able to:. Use the Ideal Gas Law to solve a gas stoichiometry problem. Gases and Stoichiometry. 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g)

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Gas Laws: Gas Stoichiometry At the conclusion of our time together, you should be able to:

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  1. Gas Laws: Gas StoichiometryAt the conclusion of our time together, you should be able to: Use the Ideal Gas Law to solve a gas stoichiometry problem.

  2. Gases and Stoichiometry 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

  3. Gases and Stoichiometry 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP? Solution 1.1 gH2O2 1 mol H2O2 1 mol O2 22.4 L O2 34 g H2O2 2 mol H2O2 1 mol O2 = 0.36 L O2at STP

  4. x 4.00 g He 1 mol He x 1 mol He 22.4 L He Gas Stoichiometry: Practice! How many grams of He are present in 8.0 L of gas at STP? 8.0 L He = 1.4 g He

  5. Gas StoichiometryTrick If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H2(g) + N2(g)  2NH3(g) 3 moles H2 + 1 mole N2  2 moles NH3 67.2 litersH2+ 22.4 liter N2 44.8 liters NH3

  6. Gas StoichiometryTrick Example How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen in a closed container at constant temperature? 3 H2(g) + N2(g)  2NH3(g) 12 L H2 2 L NH3 = L NH3 8.0 3 L H2

  7. What if the problem is NOT at STP? • 1. You will need to use PV = nRT

  8. Gas StoichiometryExample on HO How many liters of oxygen gas, at 1.00 atm and 25 oC, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 10.5 g KClO3 1 mol KClO3 3 mol O2 122.55 g KClO3 2 mol KClO3 0.13 mol O2

  9. Gas StoichiometryExample on HO How many liters of oxygen gas, at 1.00 atm and 25 oC, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) (1.0 atm) (V) (0.13 mol) (0.08206 atm*L/mol*K) (298 K) 3.2 L O2

  10. Gas Laws: Dalton, Density and Gas StoichiometryAt the conclusion of our time together, you should be able to: Explain Dalton’s Law and use it to solve a problem. Use the Ideal Gas Law to solve a gas density problem. Use the Ideal Gas Law to solve a gas stoichiometry problem.

  11. Lower density High density GAS DENSITY 22.4 L of ANY gas AT STP = 1 mole

  12. Gas Density … so at STP…

  13. Density and the Ideal Gas Law Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically: M = Molar Mass P = Pressure R = Gas Constant T = Temperature in Kelvins

  14. Gas Stoichiometry #4 How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 50.0 g KClO3 1 mol KClO3 3 mol O2 = “n” mol O2 122.55 g KClO3 2 mol KClO3 = 0.612 mol O2 = 16.7 L

  15. Try this one! How many L of O2 are needed to react 28.0 g NH3 at24°C and 0.950 atm? 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

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