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 + O2 2CO2 2 molecules 1 molecule 2 molecules 2 mol 1 mol 2 mol 2 volumes 1 volume 2 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, C3H8, 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. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(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 C3H8 x 5 L O2 / 1 L C3H8 = 1.75 L O2 • 0.350 L C3H8 x 3 L CO2 / 1 L C3H8 = 1.05 L CO2

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 Bmass 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, CaCO3, 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 CaCO3(s)  CaO(s) + CO2(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 CO2 at STP = 5.00 L • Unknown: mass of CaCO3 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 CO2 • n = (1 atm)(5.00 L)/(0.0821 L·atm/mol·K)(273 K) • n = 0.223 mol CO2 • Now that we have mol CO2, we can find mass.

11. Sample Problem 11–8 • 0.223 mol CO2 x 1 mol CaCO3 = 0.223 mol CaCO3 • 0.223 mol CaCO3 x 100.09 g CaCO3 = 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 CO2 1 mol CaCO3

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. WO3(s) + 3H2(g)  W(s) + 3H2O(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 WO3: 875 g, P of H2: 0.980 atm, T of H2: 308 K • Unknown: V of H2 in L at known conditions

13. Sample Problem 11–9 • 875 g WO3 x 1 mol WO3 = 3.77 mol WO3 • 3.77 mol WO3 x 3 mol H2 = 11.3 mol H2 • PV = nRT  V = nRT/P • V = (11.3 mol)(0.0821)(308 K) = 292 L H2 231.84 g 1 mol WO3 0.980 atm