# Stoichiometry of gases

## Presentation on theme: "Stoichiometry of gases"— Presentation transcript:

Stoichiometry 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.

Volume – Volume Calculations
For gaseous reactants and products: Coefficients can indicate volume ratios as well Only for gases 2 CO(g) O2 (g) CO2 (g) 2 molecules 1 molecules 2 molecules 2 moles 1 moles 2 moles 2 volumes 1 volumes 2 volumes Possible Volume Ratios: 2 volumes CO 2 volumes CO2 or 2 volumes CO2 2 volumes CO 2 volumes CO 1 volumes O2 or 1 volumes O2 2 volumes CO

Sample Problem 1 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 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) 5 3 Volume 0.350 L 1.75 L 1.05 L 1 5 Unlike mass : at constant temperature and pressure, volume can be determined from volume Only for GASES!!!

Volume-Mass and Mass-Volume Calculations
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

Sample 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 g mass x g/mol 1 0.223 mol 0.223 mol moles 1 ** find moles of CO2 using PV=nRT PV = nRT ** Use the value you get for moles in stoichiometry. RT RT PV (1 atm)(5.00 L) ** Can only do this for gases!!!! n = = RT L*atm mol*K ( )(273 K) = o.223 mol CO2

Sample 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) mass 875 g ÷ g/mol 3 moles 3.77 mol 11.3 mol 1 ** find moles of H2 using stoichiometry first PV = nRT ** Use PV=nRT to find the volume P P L*atm mol*K nRT (11.3 mol)(o.o )(308 K) = V = P ( atm) = 292 L

Application 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. Fuel By mass By volume  % fuel by mass Gasoline 14.7 : 1 - 6.8% Natural Gas 17.2 : 1 9.7  : 1 5.8% Propane(LP) 15.5 : 1 23.9 : 1 6.45% Ethanol 9 : 1 11.1% Methanol 6.4 : 1 15.6% Hydrogen 34 : 1 2.39 : 1 2.9% Diesel 14.6 : 1 0.094 : 1