Presentation on theme: "Gas Stoichiometry Review. Gas Stoichiometry Many chemical reactions involve gases as a reactant or a product Gas Stoichiometry – the procedure for calculating."— Presentation transcript:
Gas Stoichiometry Review
Gas Stoichiometry Many chemical reactions involve gases as a reactant or a product Gas Stoichiometry – the procedure for calculating the volume of gases as products or reactants Gases have a molar volume (L/mole) rather than concentration. This is the conversion factor used to convert (litres of gas) to (moles of gas) The Ideal Gas Law (PV = nRT) may also be required to: A) find the number of moles of reactant B) Find the V, P, or T of the product
The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies L.
What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = K P = atmo n = 49.8 g x 1 mole HCl g HCl = 1.37 mole V = 1 atmo 1.37 mole x x K Latmo molK V = 30.6 L
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atmo and 18 o C is heated to 85 o C at constant volume. What is the final pressure of argon in the lightbulb (in atmo)? PV = nRT n, V and R are constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atmo x 358 K 291 K = 1.48 atmo
Example Hydrogen gas is produced when sodium metal is added to water. What mass of sodium is necessary to produce 20.0L of hydrogen at STP? m? 20.0L 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2(g) 20.0L x 1 mole = mole H L mole H 2 x 2 mole Na = mole Na 1 mole H 2 THEN mole Na x g Na = g Na 1 mole = 41.0 g Na
Example What volume of ammonia at 450kPa and 80 o C can be obtained from the complete reaction of 7.5kg of hydrogen with nitrogen? 7500 g V ? 2N 2(g) + 3H 2(g) 2NH 3(g) 7500 g x 1 mole H2 = mole H g mole H 2 x 2 mole NH 3 = mole NH 3 3 mole H 2 Also: 450 kPa x 1 atmo = atmo and 80 o C = K kPa PV = nRT V = nRT P V = ( mole)( Latmo / moleK )(353.15K) ( atmo) = L 1.6 x 10 4 L of NH 3(g) If the conditions are not STP, the molar volume cannot be used! You must use the ideal gas law to find the gas values using moles determined from stoichiometry
Gas Stoichiometry Summary 1. Write a balanced chemical equation and list the measurements, unknown quantity symbol, and conversion factors for the measured and required substances. 2. Convert the measured quantity to moles using the appropriate conversion factor 3. Calculate the amount of moles of the required substance using the mole ratio from the balanced chemical equation. 4. Convert the calculated moles to the final quantity requested using the appropriate conversion factor.
Gas Stoichiometry Be careful! You CANNOT always use 1 mole/ L
Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mole C 6 H 12 O 6 mole CO 2 V CO g C 6 H 12 O 6 1 mole C 6 H 12 O g C 6 H 12 O 6 x 6 mole CO 2 1 mole C 6 H 12 O mole x = mole CO 2 V = nRT P mole x x K Latm molK 1.00 atm = = 4.76 L = mole C 6 H 12 O 6