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Gas Stoichiometry Review

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.

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Gas Stoichiometry Review

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  1. Gas Stoichiometry Review

  2. 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

  3. The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.41410 L. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) PV = nRT R = 0.082057 L • atm / (mol • K)

  4. 1 mole HCl V = n = 49.8 g x = 1.37 mole 36.45 g HCl 1.37 mole x 0.0821 x 273.15 K V = 1 atmo nRT L•atmo P mol•K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1.000 atmo PV = nRT V = 30.6 L

  5. P1 = 1.20 atm P2 = ? T1 = 291 K T2 = 358 K = P1 P2 T2 358 K T1 T1 T2 291 K = 1.20 atmo x P2 = P1 x 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 oC is heated to 85 oC at constant volume. What is the final pressure of argon in the lightbulb (in atmo)? n, V and Rare constant PV = nRT = 1.48 atmo

  6. 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) + 2H2O (l)  2NaOH(aq) + H2(g) 20.0L x 1 mole = 0.8922955 mole H2 22.41410 L 0.8922955 mole H2 x 2 mole Na = 1.7845909 mole Na 1 mole H2 THEN 1.7845909 mole Na x 22.9898 g Na = 41.027388 g Na 1 mole = 41.0 g Na

  7. Example 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 • What volume of ammonia at 450kPa and 80oC can be obtained from the complete reaction of 7.5kg of hydrogen with nitrogen? 7500 g V ? 2N2(g) + 3H2(g)  2NH3(g) 7500 g x 1 mole H2 = 3720.3488 mole H2 2.01594 g 3720.3488 mole H2 x 2 mole NH3 = 2480.2325 mole NH3 3 mole H2 Also: 450 kPa x 1 atmo = 4.4411547 atmo and 80oC = 353.15 K 101.325 kPa PV = nRT  V = nRT P V = (2480.2325mole)(0.08205784Latmo/moleK)(353.15K) (4.4411547 atmo) = 16183.616L  1.6 x 104 L of NH3(g)

  8. Gas Stoichiometry Summary • Write a balanced chemical equation and list the measurements, unknown quantity symbol, and conversion factors for the measured and required substances. • Convert the measured quantity to moles using the appropriate conversion factor • Calculate the amount of moles of the required substance using the mole ratio from the balanced chemical equation. • Convert the calculated moles to the final quantity requested using the appropriate conversion factor.

  9. Gas Stoichiometry Be careful! You CANNOT always use 1 mole/22.414 L

  10. What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) 6 mole CO2 g C6H12O6 mole C6H12O6 mole CO2V CO2 0.03108 mole x 1 mole C6H12O6 1 mole C6H12O6 x 180.16 g C6H12O6 L•atm mol•K nRT 0.187 mole x 0.0821 x 310.15 K = P 1.00 atm Gas Stoichiometry 5.60 g C6H12O6 = 0.03108 mole C6H12O6 = 0.187 mole CO2 V = = 4.76 L

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