Presentation on theme: "1.Use a chemical equation to specify volume ratio for gaseous reactants or products, or both. 2.Use volume ratios and the gas laws to calculate volumes,"— Presentation transcript:
1.Use a chemical equation to specify volume ratio for gaseous reactants or products, or both. 2.Use volume ratios and the gas laws to calculate volumes, masses, or molar amounts of gaseous reactants or products.
For gaseous reactants and products: Coefficients can indicate volume ratios as well Only for gases 2 CO(g) + O 2 (g) 2 CO 2 (g) 2 molecules 1 molecules 2 moles1 moles2 moles 2 volumes1 volumes2 volumes Possible Volume Ratios: 2 volumes CO 2 volumes CO 2 or 2 volumes CO 2 2 volumes CO 1 volumes O 2 or 1 volumes O 2 2 volumes CO
Propane, C3H8, is a gas that is sometimes used as a fuel for cooking and heating. The complete combustion of propane occurs according to the equation below. (a) What will be the volume, in liters, of oxygen required for the complete combustion of 0.350 L of propane? (b) What will be the volume of carbon dioxide produced in the reaction? Assume constant temperature and pressure. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) Volume Unlike mass : at constant temperature and pressure, volume can be determined from volume 0.350 L Only for GASES!!! 5 1 1.75 L 3 5 1.05 L
Problem type 1: gas volume A moles A moles B mass of B Problem type 2: mass A moles A moles B gas volume B
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. How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at STP. CaCO 3 (s) CaO (s) + CO 2 (g) PV = nRT mass moles 0.223 mol 22.3 g n = RT PV RT = (1 atm)(5.00 L) (0.0821 )(273 K) L*atm mol*K =o.223 mol CO 2 1 1 0.223 mol x 100.09 g/mol ** find moles of CO 2 using PV=nRT ** Use the value you get for moles in stoichiometry. ** Can only do this for gases!!!!
Tungsten, W, a metal used in light-bulb filaments, is produced industrially by the reaction of tungsten oxide with hydrogen. How many liters of hydrogen gas at 35 o C and 0.0980 atm are needed to react completely with 875 g of tungsten oxide. WO 3 (s) + 3 H 2 (g) W (s) + 3H 2 O(g) mass moles 875 g ÷ 231.84 g/mol 3.77 mol 3 1 11.3 mol ** find moles of H 2 using stoichiometry first ** Use PV=nRT to find the volume PV = nRT PP V = nRT P = (11.3 mol)(o.o821 )(308 K) (0.0980 atm) L*atm mol*K =292 L
Gasoline engines are run at stoichiometric air-to-fuel ratio, because gasoline is so volatile that mixes properly w/ the air. Diesel engines, in contrast, run lean, with more air available than simple stoichiometry would require. Diesel fuel is heavier and does not burn immediately to give gaseous products. Thus, it would form soot (black smoke) at stoichiometric ratio. FuelBy mass By volume % fuel by mass Gasoline14.7 : 1-6.8% Natural Gas 17.2 : 19.7 : 15.8% Propane(L P) 15.5 : 123.9 : 16.45% Ethanol9 : 1-11.1% Methanol6.4 : 1-15.6% Hydrogen34 : 12.39 : 12.9% Diesel14.6 : 10.094 : 16.8%