1. Ch. 10 - Gases III. Gas Stoichiometry at Non-STP Conditions and the last gas laws!

2. A. Gas Stoichiometry b Moles  Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas law b Non- STP 1. Given liters of gas?  start with ideal gas law, then stoich 2. Looking for liters of gas?  start with stoich conv, then ideal gas law

3. 1 mol CaCO 3 100.09g CaCO 3 B. Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 0.0525mol CO 2 CaCO 3  CaO + CO 2 1 mol CO 2 1 mol CaCO 3 5.25 g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

4. WORK: PV = nRT (103 kPa)V = (.0525 mol) x (8.315 L  kPa/mol  K ) (298K) V = 1.26 L CO 2 B. Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 0.0525 mol T = 25°C = 298 K R = 8.315 L  kPa/mol  K

5. WORK: PV = nRT (97.3 kPa) (15.0 L)= n (8.315 L  kPa/mol  K)(294K) n = 0.597 mol O 2 B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 L  kPa/mol  K Given liters: Start with Ideal Gas Law and calculate moles of O 2.

6. 2 mol Al 2 O 3 3 mol O 2 B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 0.597 mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O 3 101.96 g Al 2 O 3 1 mol Al 2 O 3 15.0L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.

7. C. Molar Mass b The equation: PV = nRT can be changed to calculate molar mass of an unknown gas. M = molar mass m= mass SO: = mRT PV M

8. C. Molar Mass Example b Find the molar mass of 0.1968 L of an unknown gas that has a mass of 0.3916 g and was collected at 22 o C and 735 mm Hg.

9. DensityDensity b From the previous equation: Since D=m/v: D= P D= P RT RT M

10. Example Problems b Find the density of hydrogen gas at STP. D = M V D = 2.0 g/mol 22.4 L/mol =.089 g/L See red/white/blue periodic table for densities of the gases

11. D. Dalton’s Law b The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 +... When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor.

12. D. Dalton’s Law & Water Displacement P total = P gas + P water vapor Measured at several temps. (see chart)

13. GIVEN: P gas = ? P total = 742.0 mm Hg P H2O = 42.18 mm Hg WORK: P total = P gas + P H2O 742.0 mm = P gas +42.18 mm P gas = 699.8 mm Hg b A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 mm Hg. What is the partial pressure of the dry gas? Look up water-vapor pressure on the chart for 35.0°C. Sig Figs: Round to least number of decimal places. D. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

14. GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 19.827 mmHg WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.64 kPa P H2 = 91.8 kPa D. Dalton’s Law b Hydrogen gas is collected over water at 22°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure and convert for 22°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

15. D. Graham’s Law b Speed of diffusion At the same temp & KE, heavier molecules move more slowly.  Larger M  smaller v (velocity)

16. D. Graham’s Law

17. b Determine the relative rate of diffusion for krypton and bromine. Kr diffuses 1.381 times faster than Br 2. B. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.

18. b A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? D. Graham’s Law Put the gas with the unknown speed as “Gas A”.

19. Root Mean Square Velocity b In chemistry, the root mean square velocity is defined as the square root of the average velocity- squared of the molecules in a gas. b R =ideal gas constant (8.314 J/(mol ⋅ K)) b T is the temp of the gas in kelvins b M is the molar mass of the compound in kilograms per mole.

20. b An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. B. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0. Square both sides to get rid of the square root sign.