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Gas Stoichiometry

Gas Stoichiometry. Relationships. 2 H 2 + O 2  2 H 2 O 2 molecules + 1 molecule yields 2 molecules 2 moles + 1 mole yields 2 moles 4 atoms + 2 atoms yields 6 atoms

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Gas Stoichiometry

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  1. Gas Stoichiometry

  2. Relationships • 2 H2 + O2 2 H2O • 2 molecules + 1 molecule yields 2 molecules • 2 moles + 1 mole yields 2 moles • 4 atoms + 2 atoms yields 6 atoms • 4 grams + 32 grams yields 36 grams • 2[6.02 x 1023] + 1[6.02 x 1023] yields 2[6.02 x 1023] particles particles particles

  3. Recalling Avagadro’s Principle • equal volumes of gases [at the same temp. and press.] contain equal numbers of molecules, • Gaseous reactions occuring at the same temperature and pressure have a volumetric relationship 2 H2 + O2 2 H2O • 2 volumes H2+ 1 volume O22 volumes H2O vapor

  4. Joseph Gay-Lussac • First chemist to apply this idea to chemical reactions involving gases. • Law of Combining Gas Volumes • [At the same temp. and press.] the volumes of reacting gases and their gaseous product[s] are expressed in small whole number ratios. • coefficients

  5. Using Gay-Lussac’s idea, Avagadro was able to identify the 7 diatomic elements. • At S.T.P. • contains 1 mole • 6.02 x 1023 chemical units of any gas • O mass should equal 16 g • O mass actually equals 32 g • Oxygen must be diatomic (O2) 22.4 L

  6. Since molar ratios are the same as volume ratios - the liter volumes can be used in the same manner as the molar ratios in calculations!Have ratio will be the same forWantmole ratios and volume ratios (when all of the substances are gases).

  7. Volume - Volume Calculations • FACT: Air is 20.9 % oxygen • Based on the equation • 2C8H18[g] + 25O2[g]  16CO2[g] +18H2O[g] • A. How many liters of AIR must enter the carburetor to complete the combustion of 30.0 L of octane, C8H18? • B. How many liters of CO2 are formed? • Assume all measurements are at the same temp. and press.

  8. 2C8H18[g] + 25O2[g]  16CO2[g] +18H2O[g] • 30 L x • 30 L C8H18 | 25 L O2 = 375 L O2 • | 2 L C8H18 • Since air is 20.9 % [.209] oxygen • 375 L O2 = [.209]x • 1794 L air = x

  9. 2C8H18[g] + 25O2[g]  16CO2[g] +18H2O[g] • 30.0 L x • 30 L C8H18 | 16 L CO2 = 240 L CO2 • | 2 L C8H18

  10. Mass - Volume Calculations • How many grams of CaCO3 must decompose to produce 5.0 L of CO2 at S.T.P.? • CaCO3[s]  CaO[s] + CO2[g] • x 5 L • *CaCO3 is chalk, a solid. • Gay-Lussac’s Law applies only for chemical reactions in which all components are gaseous.!!!!!!!!!!!!!!!

  11. What do we know?: • * 1 mole of CO2 weighs 44 g • * 1 mole of CaCO3 weighs 100 g • * 1 mole of CO2 occupies 22.4 L at S.T.P.

  12. Assuming S.T.P. • CaCO3[s]  CO2[g] +CaO[s] • x 5 L • 5 L CO2 | 1 mol CaCO3 | 100 g CaCO3 | 1 mol CO2 = 22.3 g Ca CO3 | 1 mol CO2 | 1 mol CaCO3| 22.4 L • STOICHIOMETRYMOLAR VOLUME STEP

  13. Try this one! • What volume of O2, collected by water displacement, at 20oC and 750 Torr, can be obtained by the decomposition of 55.0 g of KClO3? 2KClO3 2 KCl + 3O2 55 g x

  14. Strategy 2KClO3 2 KCl + 3O2 55 g x • First solve as if at S.T.P. 55g KClO3 | 1 mol KClO3 | 3 mol O2 | 22.4 L O2 = 15.1 L @ S.T.P. | 122.5 g KClO3 | 2 mol KClO3 | 1 mol O2

  15. Strategy • Now change to the conditions in the question. • 750 Torr - 17.5 Torr = 732.5 Torr Vapor pressure Pressure of the dry gas of water at 20oC

  16. Strategy • Using the Combined Gas Law to find the new volume. • V1 = 15.1 L V2 = x • T1 = 273 K T2 = 293 K • P1 = 760 Torr P2 = 732.5 Torr V1P1 = V2P2 T1 T2 [15.1 l][760 Torr] = x[732.5 Torr] 273 K 293 K • x = 16.9 L

  17. With Limiting Reactants • What volume of CO2 is produced at S.T.P. from 4.14 g Ca3[PO4]2 and 1.20 g SiO2? • 2Ca3[PO4]2[cr] + 6SiO2[cr] 10C[amor] + 6CaSiO3[cr] + 10CO2[g] + P4[cr] 1.34 x 10-2 mol 2.0 x 10-2 mol x • Determine the L.R. • x = 1.5 L, using Ca3[PO4]2 • x = .76 L, using SiO2 , therefore SiO2 is the L.R.

  18. And the answer is… • 2.0 x 10-2 mol SiO2 | 10 mol CO2 | 22.4 L CO2 = .747 L CO2 | 6 mol SiO2 | 1 mol CO2

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