1. Section 11–3: Stoichiometry of Gases Coach Kelsoe Chemistry Pages 347–350

2. Section 11–3 Objectives Explain how Gay-Lussac’s law and Avogadro’s law apply to the volumes of gases in chemical reactions. Use a chemical equation to specify volume ratios for gaseous reactants or products, or both. Use volume ratios and the gas laws to calculate volumes, masses, or molar amounts of gaseous reactants or products.

3. Stoichiometry of Gases You can apply the discoveries of Gay-Lussac and Avogadro to calculate the stoichiometry of reactions involving gases. For gaseous reactants or products, the coefficients in chemical equations not only indicate molar amounts and mole ratios but also reveal volume ratios. For example: 2CO + O 2  2CO 2 2 molecules1 molecule2 molecules 2 mol1 mol2 mol 2 volumes1 volume2 volumes

4. Stoichiometry of Gases So just as we could come up with molar ratios when we did stoichiometry in Chapter 9, we can also come up with volume ratios. Volumes can be compared in this way only if all are measured at the same temperature and pressure.

5. Volume–Volume Calculations Suppose the volume of a gas involved in a reaction is known and you need to find the volume of another gaseous reactant or product, assuming both reactant and product exist under the same conditions. Use volume ratios in exactly the same way you would use mole ratios.

6. Sample Problem 11–7 Propane, C 3 H 8, is a gas that is sometimes used as a fuel for cooking or heating. The complete combustion of propane occurs according to the following equation. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) What will be the volume, in liters, of oxygen required for the complete combustion of 0.350 L of propane? Carbon dioxide? 0.350 L C 3 H 8 x 5 L O 2 / 1 L C 3 H 8 = 1.75 L O 2 0.350 L C 3 H 8 x 3 L CO 2 / 1 L C 3 H 8 = 1.05 L CO 2

7. Volume–Mass and Mass– Volume Calculations Stoichiometric calculations may involve both masses and gas volumes. Sometimes the volume of a reactant or product is given and the mass of a second gaseous substance is unknown. In other gases, a mass amount may be known and a volume may be unknown. The calculations require routes such as the following: gas volume A  moles A  moles B  mass B mass A  moles A  moles B  gas volume B

8. Volume-Mass and Mass-Volume Calculations To find the unknown in cases like these, you must know the condition under which both the known and unknown gas volumes have been measured. The ideal gas law is useful for calculating values at standard and nonstandard conditions.

9. Sample Problem 11–8 Calcium carbonate, CaCO 3, also known as limestone, can be heated to produce calcium oxide (lime), an industrial chemical with a wide variety of uses. The balanced equation for the reaction follows CaCO 3 (s)  CaO(s) + CO 2 (g) How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP?

10. Sample Problem 11–8 Given: balanced equation, desired volume of CO 2 at STP = 5.00 L Unknown: mass of CaCO 3 in grams STP tells us temperature and pressure and we know the volume of carbon dioxide (5.00 L) PV = nRT  n = PV/RT for CO 2 n = (1 atm)(5.00 L)/(0.0821 L·atm/mol·K)(273 K) n = 0.223 mol CO 2 Now that we have mol CO 2, we can find mass.

11. Sample Problem 11–8 0.223 mol CO 2 x 1 mol CaCO 3 = 0.223 mol CaCO 3 0.223 mol CaCO 3 x 100.09 g CaCO 3 = 22.3 g 22.3 grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP. 1 mol CO 2 1 mol CaCO 3

12. Sample Problem 11–9 Tungsten, W, a metal used in light-bulb filaments, is produced industrially by the reaction of tungsten oxide with hydrogen. WO 3 (s) + 3H 2 (g)  W(s) + 3H 2 O(l) How many liters of hydrogen gas at 35°C and 0.980 atm are needed to react completely with 875 g of tungsten oxide? Given: balanced chemical equation, mass of WO 3 : 875 g, P of H 2 : 0.980 atm, T of H 2 : 308 K Unknown: V of H 2 in L at known conditions

13. Sample Problem 11–9 875 g WO 3 x 1 mol WO 3 = 3.77 mol WO 3 3.77 mol WO 3 x 3 mol H 2 = 11.3 mol H 2 PV = nRT  V = nRT/P V = (11.3 mol)(0.0821)(308 K) = 292 L H 2 231.84 g 1 mol WO 3 0.980 atm