Extra Practice Mixed Gas Law Problems Answers
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Mixed Extra Gas Law Practice Problems (Ideal Gas, Dalton’s Law of Partial Pressures, Graham’s Law) 1. Dry ice is carbon dioxide in the solid state. 1.28 grams of dry ice is placed in a 5.00 L chamber that is maintained at 35.1oC. What is the pressure in the chamber after all of the dry ice has sublimed? ππ = ππ π 1.28 π πΆπ! 1 ππππ πΆπ! π₯ = 0.29π849 πππππ πΆπ! 1 44.009 π πΆπ! P=? V = 5.00 L (3 significant figures) n = 0.290849 moles (3 significant figures) R = 0.0821 L*atm/mole*K (infinite significant figures) T = 35.1oC = 308.1 K (4 significant figures) 0.290849 πππππ ππ π π = = π 0.0821 πΏ − ππ‘π ππππ − πΎ 308.1 πΎ 5.00 πΏ = 1.47 ππ‘π If you used a different R, then the answers are: 1120 torr 1120 mm Hg 149 kPa 2. A sample of chlorine gas is loaded into a 0.25 L bottle at standard temperature of pressure. How many moles of bromine gas are in the container? How many grams? At STP 1 mole = 22.4 L Molar Mass of Chlorine (remember, it is a diatomic) = 70.906 g/mole Use factor label 0.25 πΏ 1 ππππ πΆπ! π₯ = 0.011 πππππ πΆπ! 1 22.4 πΏ πΆπ! 0.25 πΏ 1 ππππ πΆπ! 70.906 π πΆπ! π₯ π₯ = 0.79 π πΆπ! 1 22.4 πΏ πΆπ! 1 ππππ πΆπ! OR – use the ideal gas law, PV = nRT P = 1 atm (infinite significant figures) V = 0.25 L (2 significant figures) R = 0.0821 L*atm/mole*K (infinite significant figures) T = 273 K (infinite significant figures) π = ππ 1 ππ‘π (0.25 πΏ) = = 0.011 πππππ πΆπ! πΏ − ππ‘π π π 0.0821 (273 πΎ) ππππ − πΎ 0.011 ππππ πΆπ! 70.906 π πΆπ! π₯ = 0.78 π πΆπ! 1 1 ππππ πΆπ! 3. What is the volume of 92.4 grams of chlorine gas that is at a temperature of 46oC and a pressure of 694 mmHg? P = 694 mmHg (3 significant figures) V=? R = 62.4 L*mm Hg/mole*K (infinite significant figures) T = 46oC = 319 K (3 significant figures) n – need to calculate 92.4 πππππ πΆπ! 1 ππππ πΆπ! π₯ = 1.3π3133 πππππ πΆπ! − ππππππ ππ’ππππ ππ πππ π‘ π ππππππππππ‘ ππππ’ππ 1 70.906 π πΆπ! πΏ − πππ»π 1.303133 πππππ 62.4 (319 πΎ) ππ π ππππ − πΎ π = = = 37.4 πΏ π 694 πππ»π 4. What is the molar mass of a gas that effuses 3.7 times faster than Krypton? π ππ‘π! π! = π ππ‘π! π! A is the unknown gas (it effuses faster) B is Krypton (it effuses slower) π ππ‘π! = 3.7 π ππ‘π!" π π π ππ‘π! = 3.7 π ππ‘π!" π ππ‘π! π!" = = 3.7 π ππ‘π!" π! Square both sides 13.69 = π!" π! Rearrange and solve for mass of the unknown π 84.80 π!" π ππππ π! = = = 6.19 13.69 13.69 ππππ 5. Carbon dioxide is collected over water at a temperature of 18oC. The pressure of water vapor at 18oC is 2.20 kPa. If the pressure of the gas collected is 1.3 atm, what is the pressure of the dry gas alone? π!"#$% = 1.3 ππ‘π π!"#$% !"#$% = 2.20 πππ = 2.20 πππ 1 ππ‘π π₯ = 0.021π1767 ππ‘π 1 101.3 πππ π!"#$%& !"#$"!% = π!"!#$ − π!"#$% !"#$% = 1.3 ππ‘π − 0.02171767 ππ‘π = 1.3 ππ‘π 6. What diffuses more quickly: tetracarbon decahydride or iodine? By how much? C4H10 versus I2 Molar mass of C4H10 = 58.123 g/mole Molar mass of I2 = 253.808 g/mole Butane will effuse more quickly because it has a smaller molar mass π ππ‘π! π! = π ππ‘π! π! A – butane B – iodine π 253.808 π ππ‘π! ππππ = π = 2.09 π ππ‘π! 58.123 ππππ 7. What is the molar mass of a gas that effuses 4.2 times faster than bromine? π ππ‘π! π! = π ππ‘π! π! A is the unknown gas (it effuses faster) B is Bromine (it effuses slower) π ππ‘π! = 4.2 π ππ‘π!"! π π π ππ‘π! = 4.2 π ππ‘π!"! π ππ‘π! π!" = = 4.2 π ππ‘π!"! π! Square both sides 17.64 = Rearrange and solve for mass of the unknown π!"! π! π! = π!"! 17.64 = π ππππ = 9.06 π 17.64 ππππ 159.808 8. A mixture of gases contains oxygen, water vapor, carbon dioxide, and argon. If the partial pressure of oxygen is 3.7 atm, the partial pressure of the water vapor is 382 mmHg, the partial pressure of the carbon dioxide is 234 kPa, and the partial pressure of the argon is 143 mmHg, what is the total pressure? π!! = 3.7 ππ‘π π!! ! = 382 ππ π»π = 382 ππ π»π 1 ππ‘π π₯ = 0.50π63 ππ‘π 1 760 ππ π»π π!"! = 234 πππ = π!" = 143 ππ π»π = 234 πππ 1 ππ‘π π₯ = 2.3π997 ππ‘π 1 101.3 πππ 143 ππ π»π 1 ππ‘π π₯ = 0.18π157 ππ‘π 1 760 ππ π»π π!"!#$ = π!! + π!! ! + π!"! + π!" = 3.7 ππ‘π + 0.50π63 ππ‘π + 2.3π997 πππ + 0.18π157 ππ‘π = 6.7 ππ‘π 9. What is the temperature of 83.28 grams of carbon dioxide at a pressure of 1830 torr contained in a 3.82 L container? P = 1830 torr (3 significant figures) V = 3.82 L (3 significant figures) R = 62.4 L*torr/mole*K (infinite significant figures) T=? n – need to calculate 83.28 πππππ πΆπ! 1 ππππ πΆπ! π₯ = 1.89π340 πππππ πΆπ! − ππππππ ππ’ππππ ππ πππ π‘ π ππππππππππ‘ ππππ’ππ 1 44.009 π πΆπ! π = ππ 1830 π‘πππ (3.82 πΏ) = = 59.2 πΏ πΏ − π‘πππ ππ (1.892340 πππππ ) 62.4 ππππ − πΎ
