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p.337 2. using this you can do stoichiometry problems with gases!

p.337 2. using this you can do stoichiometry problems with gases! vol of gas in L 1 mol 22.4L Make your life easier…remember this:

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p.337 2. using this you can do stoichiometry problems with gases!

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  1. p.337 • 2. using this you can do stoichiometry problems with gases! vol of gas in L 1 mol 22.4L • Make your life easier…remember this: For gaseous reactants or products the coefficients in chemical equations indicate molar amts, mole ratios AND volume ratios!!

  2. 2CO + O2 → 2 CO2 • 2 mol 1 mol 2 mol • 2 mL 1 mL 2 mL • 2 L 1 L 2 L • So, with gases you can do volume-volume calculations.

  3. G. Ideal Gas Law--- shows relationship between P, T, V, & number of moles of a gas • PV = nRT n is the number of moles R is the ideal gas constant

  4. Which R you use depends on which units the pressure is measured in! p • p.342 • R= 62.4 if using mm Hg • R = 0.0821 if using atm • R = 8.314 if using Pa • R = 8.314 if using kPa

  5. What can you do with this? • Once you have the number of moles of a gas you can calculate mass! • If you have an unknown gas you can find the molar mass (M)! M = g/mol • M = mass in g M = m # of moles n • AND…

  6. do vol:mass calculations • AND • If you know the mass & volume then you can calculate density! d = m/v STP

  7. What mass solid magnesium is required to react w/250 mL carbon dioxide 1.5 atm and 77oC to produce solid magnesium oxide and solid carbon? 2 Mg (s) + CO2 (g) 2 MgO (s) + C (s) ? g Mg V = 250 mL 0.25 L T = 77oC 350 K Use PV = nRT to find number of moles then convert moles to grams!! Use moles of CO2 & convert to grams of magnesium! .013 mol CO2 2 mol Mg 24.305 g Mg = .63g Mg 1 mol CO2 1 mol Mg

  8. Apply the Gas Law • The pressure shown on a tire gauge doubles as twice the volume of air is added at the same temperature. • A balloon over the mouth of a bottle containing air begins to inflate as it stands in the sunlight. • An automobile piston compresses gases. • An inflated raft gets softer when some of the gas is allowed to escape. Avogadro’s principle Charles’s law Boyle’s law Avogadro’s principle

  9. Application of gas laws • A balloon placed in the freezer decreases in size. • A hot air balloon takes off when burners heat the air under its open end. • When you squeeze an inflated balloon, it seems to push back harder. • A tank of helium gas will fill hundreds of balloons. • Model: When red, blue, and white Ping-Pong balls are shaken in a box, the effect is the same as if an equal number of red balls were in the box. Charles’s law Charles’s law Boyle’s law Boyle’s law Dalton’s law

  10. Gas Stoichiometry Moles  Liters of a Gas: • STP - use 22.4 L/mol • Non-STP - use ideal gas law Non-STP • Given liters of gas? • start with ideal gas law • Looking for liters of gas? • start with stoichiometry conv. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  11. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? CaCO3 CaO + CO2 5.25 g ? Lnon-STP Looking for liters: Start with stoichyand calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 100.09g CaCO3 1 mol CO2 1 mol CaCO3 = .052 mol CO2 Plug this into the Ideal Gas Law to find liters. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  12. PV = nRT • (103kPa) V = (.052) (8.314 LkPa/mol K) (298K) • Or V = nRT V = (.052) (8.314) (298K) P 103kPa • V = 1.25L

  13. Gas Stoichiometry Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid. Pres.=107.3 kPa; temp.= 88oC. Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g) P = 107.3 kPa T = 88oC *** 38.2 g Zn excess 38.2g Zn 1 mol Zn 1 mol H2 = 0.584 mol H2 65.4g Zn 1 mol Zn At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP.

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