1. Gas Stoichiometry

2. 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  4 grams + 32 grams yields 36 grams  2[6.02 x 10 23 ] + 1[6.02 x 10 23 ] yields 2[6.02 x 10 23 ] particles particles particles 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 H 2 + O 2  2 H 2 O  2 volumes H 2 + 1 volume O 2  2 volumes H 2 O 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 10 23 chemical units of any gas of any gas  O mass should equal 16 g  O mass actually equals 32 g   Oxygen must be diatomic (O 2 ) 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 for Want mole ratios and volume ratios (when all of the substances are gases).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 for Want mole 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  2C 8 H 18 [g] + 25O 2 [g]  16CO 2 [g] +18H 2 O[g]  A. How many liters of AIR must enter the carburetor to complete the combustion of 30.0 L of octane, C 8 H 18 ?  B. How many liters of CO 2 are formed?  Assume all measurements are at the same temp. and press.

8.  2C 8 H 18 [g] + 25O 2 [g]  16CO 2 [g] +18H 2 O[g]  30 L x  30 L C 8 H 18 | 25 L O 2 = 375 L O 2  | 2 L C 8 H 18  Since air is 20.9 % [.209] oxygen  375 L O 2 = [.209]x  1794 L air = x

9.  2C 8 H 18 [g] + 25O 2 [g]  16CO 2 [g] +18H 2 O[g]  30.0 L x  30 L C 8 H 18 | 16 L CO 2 = 240 L CO 2  | 2 L C 8 H 18

10. Mass - Volume Calculations  How many grams of CaCO 3 must decompose to produce 5.0 L of CO 2 at S.T.P.?  CaCO 3 [s]  CaO[s] + CO 2 [g]  x 5 L  *CaCO 3 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 CO 2 weighs 44 g  * 1 mole of CaCO 3 weighs 100 g  * 1 mole of CO 2 occupies 22.4 L at S.T.P. S.T.P.

12. Assuming S.T.P.  CaCO 3 [s]  CO 2 [g] +CaO[s]  x 5 L  5 L CO 2 | 1 mol CaCO 3 | 100 g CaCO 3 | 1 mol CO 2 = 22.3 g Ca CO 3 | 1 mol CO 2 | 1 mol CaCO 3 | 22.4 L | 1 mol CO 2 | 1 mol CaCO 3 | 22.4 L  STOICHIOMETRY MOLAR VOLUME STEP STEP

13. Try this one!  What volume of O 2, collected by water displacement, at 20 o C and 750 Torr, can be obtained by the decomposition of 55.0 g of KClO 3 ? 2KClO 3  2 KCl + 3O 2 55 g x 55 g x

14. Strategy 2KClO 3  2 KCl + 3O 2 55 g x 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 | 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 Vapor pressure Pressure of the dry gas of water at 20 o C of water at 20 o C

16. Strategy  Using the Combined Gas Law to find the new volume.  V 1 = 15.1 LV 2 = x  T 1 = 273 KT 2 = 293 K  P 1 = 760 TorrP 2 = 732.5 Torr V 1 P 1 = V 2 P 2 T 1 T 2 T 1 T 2  [15.1 l][760 Torr] = x[732.5 Torr] [15.1 l][760 Torr] = x[732.5 Torr] 273 K 293 K 273 K 293 K  x = 16.9 L

17. With Limiting Reactants  What volume of CO 2 is produced at S.T.P. from 4.14 g Ca 3 [PO 4 ] 2 and 1.20 g SiO 2 ?  2Ca 3 [PO 4 ] 2[cr] + 6SiO 2[cr]  10C [amor] + 6CaSiO 3[cr] + 10CO 2[g] + P 4[cr] 1.34 x 10-2 mol 2.0 x 10-2 mol x 1.34 x 10-2 mol 2.0 x 10-2 mol x  Determine the L.R. –x = 1.5 L, using Ca 3 [PO 4 ] 2 –x =.76 L, using SiO 2, therefore SiO 2 is the L.R.

18. And the answer is…  2.0 x 10-2 mol SiO 2 | 10 mol CO 2 | 22.4 L CO 2 =.747 L CO 2 | 6 mol SiO 2 | 1 mol CO 2 | 6 mol SiO 2 | 1 mol CO 2