Presentation on theme: "Stoichiometry of gases"— Presentation transcript:
1Stoichiometry of gases Objectives:Use a chemical equation to specify volume ratio 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.
2Volume – Volume Calculations For gaseous reactants and products:Coefficients can indicate volume ratios as wellOnly for gases2 CO(g) O2 (g) CO2 (g)2 molecules1 molecules2 molecules2 moles1 moles2 moles2 volumes1 volumes2 volumesPossible Volume Ratios:2 volumes CO2 volumes CO2or2 volumes CO22 volumes CO2 volumes CO1 volumes O2or1 volumes O22 volumes CO
3Sample Problem 1Propane, 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 L of propane?(b) What will be the volume of carbon dioxide produced in the reaction?Assume constant temperature and pressure.C3H8(g) O2(g) CO2(g) H2O(g)53Volume0.350 L1.75 L1.05 L15Unlike mass : at constant temperature and pressure, volume can be determined from volumeOnly for GASES!!!
4Volume-Mass and Mass-Volume Calculations Problem type 1:gas volume A moles A moles B mass of BProblem type 2:mass A moles A moles B gas volume B
5Sample Problem 2 CaCO3 (s) CaO (s) + CO2 (g) PV = nRT RT RT PV Calcium carbonate, CaCO3, 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.CaCO3 (s) CaO (s) CO2 (g)22.3 gmassx g/mol10.223 mol0.223 molmoles1** find moles of CO2 using PV=nRTPV = nRT** Use the value you get for moles in stoichiometry.RTRTPV(1 atm)(5.00 L)** Can only do this for gases!!!!n ==RTL*atmmol*K( )(273 K)=o.223 mol CO2
6Sample Problem 3 WO3 (s) + 3 H2 (g) W (s) + 3H2O(g) PV = nRT P P nRT 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 35oC and atm are needed to react completely with 875 g of tungsten oxide.WO3 (s) H2 (g) W (s) H2O(g)mass875 g÷ g/mol3moles3.77 mol11.3 mol1** find moles of H2 using stoichiometry firstPV = nRT** Use PV=nRT to find the volumePPL*atmmol*KnRT(11.3 mol)(o.o )(308 K)=V =P( atm)=292 L
7ApplicationGasoline 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 massBy volume % fuel by massGasoline14.7 : 1-6.8%Natural Gas17.2 : 19.7 : 15.8%Propane(LP)15.5 : 123.9 : 16.45%Ethanol9 : 111.1%Methanol6.4 : 115.6%Hydrogen34 : 12.39 : 12.9%Diesel14.6 : 10.094 : 1