1. Gas Stoichiometry Balanced chemical equations can be used to relate moles or grams of reactant(s) to products. When gases are involved, these relations can be applied to include volume.

2. Avogadro’s Law “Equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.”

3. Example of a Gas Stoichiometry Problem Airbags in automobiles contain sodium azide (NaN 3 ), potassium nitrate, and silicon dioxide. (All are solids.) 1.Upon impact, the bag is inflated by the thermal decomposition of sodium azide (NaN 3 ) to sodium metal and nitrogen gas. 2.Because sodium is toxic and very reactive, it reacts with the potassium nitrate to produce potassium oxide and sodium oxide, and (additional) nitrogen gas. 3.The metal oxides are removed by reacting with the silicon dioxide to produce alkaline silicate (glass). Question. The driver’s airbag fills to 50-60 liters. Assuming the pressure inside the airbag is 1 atm, calculate the number of grams of each solid substance needed for a 50-L airbag.

4. When acid is added to sodium bicarbonate (sodium hydrogen carbonate), NaHCO 3, the following reaction occurs: NaHCO 3 (s) + H + (aq)  Na + (aq) + H 2 CO 3 (aq) but H 2 CO 3 (aq) quickly decomposes to CO 2 + H 2 O, so the actual reaction is: NaHCO 3 (s) + H + (aq)  Na + (aq) + CO 2 (g) + H 2 O (l) All of the following experiments are performed with 2.45 M HCl and 12.75 g of NaHCO 3 at 732 mm Hg and 38 o C. Example 5.6

5. 2.45 M HCl and 12.75 g of NaHCO 3 at 732 mm Hg and 38 o C. (a) If an excess of HCl is used, what volume of CO 2 is produced? NOTE: The first solution that follows is ‘the long way around.’ After that, the solution is a shorter way that you should be able to do. The strategies are the same. Strategy Balanced chemical equation (1 NaHCO 3 = 1 CO 2 ) (in pb.) (i) (i)Find amount of CO 2 produced using stoichiometry (ii) (ii)Static conditions so use PV = nRT (note: you can’t use 22.4 L = 1 mol b/c it’s not at STP)

6. 2.45 M HCl and 12.75 g of NaHCO 3 at 732 mm Hg and 38 o C. (a) If an excess of HCl is used, what volume of CO 2 is produced? NaHCO 3 (s) + H + (aq)  Na + (aq) + CO 2 (g) + H 2 O (l)

7. 2.45 M HCl and 12.75 g of NaHCO 3 at 732 mm Hg and 38 o C. (a) If an excess of HCl is used, what volume of CO 2 is produced? NaHCO 3 (s) + H + (aq)  Na + (aq) + CO 2 (g) + H 2 O (l)

8. 2.45 M HCl and 12.75 g of NaHCO 3 at 732 mm Hg and 38 o C. (b) If NaHCO 3 is in excess, what volume of HCl is required to produce 2.65 L of CO 2 ?

9. 2.45 M HCl and 12.75 g of NaHCO 3 at 732 mm Hg and 38 o C. (c) What volume of CO 2 is produced when all of the NaHCO 3 is made to react with 50.0 mL HCl? Strategy (i) (i)This involves determining the limiting reagent (reactant). (ii) (ii)Convert n CO2

10. NaHCO 3 (s) + H + (aq)  Na + (aq) + CO 2 (g) + H 2 O (l) 2.45 M HCl and 12.75 g of NaHCO 3 at 732 mm Hg and 38 o C. (c) What volume of CO 2 is produced when all of the NaHCO 3 is made to react with 50.0 mL HCl?

11. N.B. Solving the previous problems, I repeated many steps that you wouldn’t when solving the problem. For example, the number of moles was calculated in (a) and used again in (c). On the AP exam, you (and they) wouldn’t want you to repeat calculations. Work smarter, not harder.