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Gas Stoichiometry Demonstrations

Gas Stoichiometry Demonstrations. Electrolysis of water. 2 H 2 O (l)  2 H 2(g) + O 2(g). Equal v olumes of gases that have the same temperature and pressure will have the same number of particles. Observe the demonstration. 2 H 2 O (l)  2 H 2(g) + O 2(g).

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Gas Stoichiometry Demonstrations

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  1. Gas Stoichiometry Demonstrations

  2. Electrolysis of water

  3. 2 H2O(l)  2 H2(g) + O2(g) • Equal volumes of gases that have the same temperature and pressure will have the same number of particles. • Observe the demonstration..

  4. 2 H2O(l)  2 H2(g) + O2(g) If 5.604 grams of water is decomposed, calculate the volumes of hydrogen and oxygen gases collected at a temperature of 25 oC and 1.0 atm pressure.

  5. 2 H2O(l)  2 H2(g) + O2(g) Convert to moles: 5.604 grams/18.01 gmol-1= 0.3000 moles of water Find moles of hydrogen and oxygen 2:2 ratio, so 0.3000 moles of hydrogen produced 2: 1 ratio, so 0.1500 moles of oxygen produced

  6. 2 H2O(l)  2 H2(g) + O2(g) Use PV = nRT Solve for V = nRT/P V = (0.3000 moles) (0.0821) (298) / 1.0 atm V = 7.33 liters of hydrogen 2:1 ratio between hydrogen and oxygen: V = 3.67 liters of oxygen

  7. Decomposition of Copper(II) Carbonate basic

  8. CuCO3.Cu(OH)2(s)  2 CuO(s) + H2O(l) + CO2(g) If 2.22 grams of Copper(II) carbonate basic are decomposed, calculate the volume of carbon dioxide gas that will be collected at a temperature of 25 oC and a pressure of 0.95 atm.

  9. CuCO3.Cu(OH)2(s)  2 CuO(s) + H2O(l) + CO2(g) • Convert grams to moles: 2.22 grams / 221.13 gmol-1 = 0.0100 mole The ratio is 1:1, so 0.0100 moles of CO2 will be produced. Use PV = nRT to find the volume of the gas. =

  10. CuCO3.Cu(OH)2(s)  2 CuO(s) + H2O(l) + CO2(g) • PV = nRT Solve for V = nRT/P V = (0.0100 moles) (0.0821) (298) / 0.95 atm V = 0.26 liters = 260 mLs

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