1. Stoich with Gases!

2. How do we figure out how many particles there are in a gas sample?

3. Avagadro’s Law Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules The volume occupied by one mole of a gas at STP is known as the standard molar volume of a gas and is equal to 22.4 L.

4. Example Problem 1: A chemical reaction produces 0.0680 mol of Oxygen gas. What volume in liters is occupied by this gas sample at STP? 0.0680 mol x 22.4 L = 1.52 L O 2 1 mol

5. Example Problem 2: A chemical reaction produced 98.0 mL of sulfur dioxide gas, SO 2, at STP. What was the mass in grams of the gas produced? 98.0 mL x 1 L x 1 mol x 64.07 g SO 2 = 0.28 g SO 2 1000 mL 22.4 L1 mol SO 2

6. Your Turn 1.What is the mass of 1.33 x 10 4 mL of oxygen gas at STP? 2.What is the volume of 77.0 g of nitrogen dioxide gas at STP? 3.At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas?

7. What is the mass of 1.33 x 10 4 mL of oxygen gas at STP? 1.33 x 10 4 mL x 1 L x 1 mole x 32 g = 19 grams 1000 mL 22.4 L 1 mol

8. What is the volume of 77.0 g of nitrogen dioxide gas at STP? 77.0 g NO 2 x 1 mol NO 2 x 22.4 L = 37.5 L 46 g NO 2 1 mol

9. At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas? 3 L Cl 2 x 1 mole x 70.9 g = 9.5 grams 22.4 L 1 mol

10. What would happen to the pressure of a gas in a fixed container if you added more molecules of gas? What would happen to the volume of a gas in a flexible container if you added more molecules of gas?

11. Volume, Temperature, Pressure, and number of moles/molecules are all related. Ideal Gas Law: mathematical relationship between pressure, volume, temperature, and the number of moles of a gas V = nRT P OR PV = nRT Moles Ideal Gas Constant

12. PV = nRT Ideal Gas Constant: R = 0.0821 L·atm mol·K R = 8.314 J mol·K

13. Example Problem: What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 25°C? 25 + 273 = 298 K PV = nRTso P = nRT V P = (0.5 mol)(0.0821L·atm)(298 K) = 1.22 atm 10.0 L mol·K

14. Your Turn What is the volume in liters of 0.25 mol of oxygen gas at 20°C and 0.974 atm pressure? 20 + 273 = 293 K PV = nRTso V = nRT P V = (0.25 mol)(0.0821L·atm)(293 K) = 6.17 L O 2 0.974 atm mol·K

15. Another What mass of chlorine gas, Cl 2, in grams, is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure? 27 + 273 = 300 K PV = nRTso n = PV RT n = (3.50 atm)(10.0 L) = 1.42 mol Cl 2 x 70.9 g = 101 g Cl 2 (0.0821L·atm)(300 K) 1 mol mol·K

16. Stoichiometry of Gases The coefficients in chemical equation not only indicate mole ratios, but also reveal volume ratios. 2CO(g) + O 2 (g)  2 CO 2 (g) 2 molecules 1 molecule2 molecule 2 mol 1 mol2 mol 2 volumes 1 volume2 volumes

17. Stoichiometry of Gases Propane C3H8 is used for cooking and heating. The combustion of propane occurs according to the following equation: C 3 H 8 (g) + 5O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) a) What is the volume, in liters, of oxygen required for the complete combustion of 0.350L of propane? b) What will be the volume of carbon dioxide produced?

18. Putting it all together: Calcium carbonate (limestone) can be heated to produce calcium oxide (lime). The balanced equation is CaCO 3(s)  CaO(s) + CO 2 (g) How many grams of CaCO 3 must be decomposed to produce 5.00 L of carbon dioxide gas at STP? STP: T = 273. K P = 1.00 atm PV = nRTso n = PV RT n = (1.00 atm) (5.00 L) = 0.223 mol CO 2 x 1 mol CaCO 3 x 100.08 g CaCO 3 = 22.3 g (0.0821L·atm)(273 K) 1 mol CO 2 1 mol CaCO 3 mol·K First use ideal gas law to calculate the number of moles of CO2 produced Then use mole ratio and molar mass to calculate the mass of CaCO 3 needed.

19. Putting it all together: Calcium carbonate (limestone) can be heated to produce calcium oxide (lime). The balanced equation is CaCO 3(s)  CaO(s) + CO 2 (g) How many grams of CaCO 3 must be decomposed to produce 5.00 L of carbon dioxide gas at STP? STP: T = 273 K P = 1.00 atm Method 2: at STP 1.00 mol of gas = 22.4 L of gas 5.00 L CO2 x 1.00mol CO 2 x 1 mol CaCO 3 x 100.08 g CaCO 3 = 22.3 g 22.4 L CO 2 1 mol CO 2 1 mol CaCO 3

20. One more: Tungsten, W, is produced industrially by the reaction of tungsten oxide with hydrogen WO 3(s) + 3H 2(g)  W (s) + 3H 2 O (g) How many liters of hydrogen gas at 35°C and 0.980 atm are needed to react completely with 875 g of tungsten oxide? 0.980 atm