Chemistry 101 ANSWER KEY REVIEW QUESTIONS Chapter 5 1. Determine the pressure of the gas (in mmhg) in the diagram below, given atmospheric pressure= 0.975 atm. atm = 0.975 atm h = 5 cmhg gas atm 760 mmhg x = 741 mmhg 10 mmhg x 1 cmhg = 50 mmhg = - h = 741 mmhg - 50 mmhg = 1 mmhg. A sample of oxygen gas has a volume of 6.7 L at 75 mmhg and 0 C. What is the volume of this gas at 1.30 atm and 0 C? V = V x 1 1 = 6.7 L x 75 mmhg = 0.3 L 760 mmhg 1.30 atm x 3. A 35.8 L cylinder of Argon gas is connected to and transferred into an evacuated 1875-L tank at constant temperature. If the final pressure in the tank is 71 mmhg, what must have been the original pressure (in atm) in the cylinder? = x 1 V = (71 mmhg V 1 x 760 mmhg 1875 L ) x = 49.7 atm 35.8 L 4. A 34.0-L cylinder contains 305 g of oxygen gas at C. How many grams of gas must be released to reduce the pressure in the cylinder to 1.15 atm if the temperature remains constant? Mass of gas present at 1.15 atm V n RT (1.15 atm )(34.0 L ) (0.081 )(95 K mol K mass of oxygen = 1.614 mol Mass of gas that must be removed 305 g 51.6 g = 53 g = 1.614 mol ) 3.0 g x = 51.6 g 1
5. At ST, 0.80 L of a gas weighs 0.400 g. Calculate the molar mass of this gas. moles of gas = 0.80 L x.4 L = 0.015 mol molar mass = 0.400 g = 3.0 g/mol 0.015 mol 6. Calculate the density of HBr gas in g/l at 733 mmhg and 46 C. 733 ( atm)(80.91 g/mol) M d 760 ) =.98 g/l RT (0.081Latm/molK)(319 K 7. A mixture of 4.00 g of hydrogen and 10.0 g of helium are in a 4.30-L flask at 0 C. What is the total pressure of the container and the partial pressures of each gas? mol H = 4.00 g mol = 10.0 g x.0 g = 1.98 mol x 4.00 g =.50 mol Total mol = 1.98 +.50 = 4.48 mol (4.48 mol )(0.081 Total mol K )(73 K ) = 3.4 atm V 4.30 L 1.98 mol.50 mol H 0.44 0.558 4.48 mol 4.48 mol = = (0.44)(3.4 atm) = 10.3 atm H H Tot = = (0.558)(3.4 atm) = 13. Tot
8. Life rafts and weather balloons can be inflated by the reaction shown below: CaH (s) + HO (l) Ca(OH) (aq) + H (g) How many grams of CaH are needed to produce 10.0 L of hydrogen gas at 740 mmhg and 3 C? (740 mmhg x )(10.0 L ) V 760 mmhg n H = 0.4007 mol RT (0.081 )(96 K ) mol K 0.4007 mol H CaH mol H x 4.10 g x = 8.43 g CaH 9. Nitroglycerin, an explosive compound, decomposes according to the equation below: 4 C3H5(NO3)3 (s) 1 CO (g) + 10 HO (g) + 6 N (g) + O (g) Calculate the total volume of gases produced at 1. atm and 6 C when 60 g of nitroglycerine is decomposed. 60 g NG V x 7.1 g 9 mol gas x 4 mol NG = 8.30 mol gas (8.30 mol )(0.081 )(99 K ) mol K = 170 L 1. atm 10. A 1.65-g sample of Al is reacted with excess HCl and the hydrogen produced is collected over water at 5 C at a barometric pressure of 744 mmhg. What volume of hydrogen gas is produced in this reaction? (Vapor pressure of water at 5 C is 3.8 mmhg) Al (s) + 6 HCl (aq) AlCl3 (aq) + 3 H (g) H = atm - HO = 744 mmhg - 3.8 mmhg = 70. mmhg 3 mol H 1.65 g Al x x = 0.09173 mol H 7.0 g mol Al V (0.09173 mol )(0.081 )(98 K ) mol K =.37 L 70. mmhg x 760 mmhg 3
11. Nitric oxide reacts with oxygen gas as shown below: NO (g) + O (g) NO (g) Initially NO and O are separated as shown in the diagram below. When the valve is opened the reaction quickly goes to completion. Determine the identity of the gases that remain at the end of the reaction and their partial pressure. Assume temperature remains at 5 C. Since the system is at constant temperature, the moles of gas is proportional to the product of pressure and volume as shown below: V = n RT at constant temp. V n Before reaction mol NO α (4.00 L)(0.500 atm) =.00 mol O α (.00 L)(1.00 atm) =.00 NO + O NO Initial.00.00 0 React.00 1.00.00 End 0 1.00.00 At end of reaction, total volume is 6.00 L. Therefore, mol 1.00 O α 0.167 atm V 6.00 mol.00 NO α 0.333 atm V 6.00 4
1. A 4.85-g sample of solid ammonium chloride is placed in a 1.50-L evacuated flask and heated until it decomposes, as shown below: NH4Cl (s) NH3 (g) + HCl (g) After the reaction has completed, the total pressure in the flask is measured as 4.40 atm at 0 C. Based on this information, what percent of the ammonium chloride decomposed? = + and = total NH3 HCl NH3 HCl 4.40 atm NH3.0 atm V (.0 atm )(1.50 L ) n NH3 = 0.0846 mol RT (0.081 mol K )(475 K ) NH4Cl 53.50 g 0.0846 mol NH 3 x x = 4.56 g NH4Cl reacted NH 3 4.56 g % decomposed = x100 = 93.3% 4.85 g 13. A mixture of gases containing 1.45 g of H, 60.67 g of N and.380 g of NH3 are placed in a 10.00 L container at 90 C. What is the total pressure (in atm) and partial pressure of each component in the gas mixture? mol H = 1.45 g mol N = 60.67 g mol NH =.380 g 3 x.016 g x 8.0 g x 17.04 g = 6.176 mol =.165 mol = 0.1397 mol Total mol = 6.176 +.165 + 0.1397 = 8.48 (8.48 )(0.081 Total mol K )(363 K ) = 5.8 atm V 10.00 L 6.176 mol H 0.78 H = H Tot = (0.78)(5.8 atm) = 18.4 8.48.165 mol 0.553 = Tot = (0.553)(5.8 atm) = 6.454 atm 8.48 0.1397 mol NH30.01647 NH3 = NH3 Tot = (0.01647)(5.8 atm) = 0.4164 atm 8.48 5
14. Which has a higher average speed, H at 150 K or at 375 C? 3RT 3 (8.314 J/molK)(150 K) u H = 1360 m/s M.0x10 kg/mol 3RT 3 (8.314 J/molK)(648 K) u = 010 m/s M 4.00x10 kg/mol o at 375 C has higher average speed 15. A big-league fastball travels at about 45.0 m/s. At what temperature ( C) do helium atoms have this same average speed? 3RT 3 (8.314 J/molK)(T) u = 45.0 m/s M 4.00x10 kg/mol squaring both sides, 3 (8.314 J/molK)(T) = 05 m /s 4.00x10 kg/mol T = 0.35 K T = 0.35-73.15 o = -7.83 C 16. The surface temperature of Venus is about 1050 K, and the pressure is about 75.0 earth atmospheres. Assuming that these conditions represent a Venusian ST, what is the standard molar volume of a gas on Venus? V (1.00 mol )(0.08 K )(1050 K ) = 1.15 L 75.0 atm 6
17. Gaseous iodine IF5 can be prepared by the reaction of solid iodine and gaseous fluorine: I (s) + 5 F (g) IF5 (g) A 5.00-L flask is charged with 10.0 g of I and 10.0 g of F, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete the temperature in the flask is 15 C. What is the partial pressure of IF5 in the flask? Before reaction, mol I = 10.0 g x = 0.0394 mol 53.8 g mol F = 10.0 g x = 0.63 mol 38.00 g I + 5 F IF 5 Initial 0.0394 0.63 0 React 0.0394 0.197 0.0788 End 0 0.066 0.0788 At end of reaction, Total moles = 0.066 + 0.0788 = 0.1448 mol nto talrt (0.1448 mol)(0.81 Latm/molK)(398 K) total = 0.946 atm V 5.00 L 0.0788 mol IF 0.544 5 IF = 5 IF 5 total = (0.544)(0.946 atm) = 0.515 atm 0.1448 mol 7