377 GAS LAW WORKSHEET 1 KEY 1. A sample of oxygen gas occupies a volume of 436. ml at 1.0 atm. If the temperature is held constant, what would the pressure of this gas be when the gas is compressed to 31.6 ml? V 436. ml = since T = T V = V = = 1.0 atm = 1.9 atm T T V 31.6 ml 1 1 1 1 1 1 1 1 To check with reasoning: decreased V leads to increased, so the volume ratio should be greater than one to yield the expected increased pressure.. If a gas originally occupying 6.75 L at 19.1 ºC and 76.5 torr is compressed to give a pressure of 1.6 atm at 6.35 ºC, what would the new volume be? 1 1 T 1 99.50 K 76.5 torr 1 atm 1 = 6.75 L = T1 T T1 9.36 K 1.6 atm 760 torr 5.51 L To check with reasoning: increased T leads to increased V, so the temperature ratio should be greater than one to yield the expected increase in volume. Increased leads to decreased V, so the pressure ratio should be less than one to yield the expected decreased volume. 3. Calculate the number of grams of hydrogen sulfide gas, H S, in a 3.6 L container at 3.6 ºC and 71 mmhg. g 71 mmhg ( 3.6 L ) 34.080 g M mol 1 atm = RT g = = = M RT 0.08058 ( 305.8 K) 760 mmhg 4.6 g HS 4. Calculate the density of 0.65 g of carbon dioxide at 6.3 ºC and 1.03 atm. g 1.03 atm 44.010 g g M mol = RT = = = 1.84 g/l CO M V RT 0.08058 ( 99.47 K)
378 Chapter 10 Worksheet Keys CHEMISTRY 151 COMBINED GAS LAW KEY 1. A McLeod gauge is an instrument used to measure extremely low pressures. Assume that a 50.0 ml sample of gas from a low pressure system is compressed in a McLeod gauge to a volume of 0.055 ml, where the pressure of the sample is 0.0355 atm. What was the original pressure of the gas in the system? V 1 = 50.0 ml V = 0.055 ml = 0.0355 atm 1 =? 0.055 ml 1V 1 = 1 = = 0.0355 atm = V1 50.0 ml 6 7.46 x 10 atm. A balloon containing 5.0 dm 3 of gas at 14 C and 100.0 ka rises to an altitude of 000 m, where the temperature is 0 C. The pressure of the gas in the balloon is now 79.0 ka. What is the volume of gas in the balloon at this altitude? V 1 = 5.0 dm 3 T 1 = 14 ºC + 73.15 = 87 K 1 = 100.0 ka T = 0 ºC + 73.15 = 93 K = 79.0 ka V =? T 93 K 100.0 ka 1 1 1 3 1 = 5.0 dm = T1 T T1 87 K 79.0 ka 3 6.5 dm 3. A cylinder contains carbon dioxide gas at 6.50 atm. Enough carbon dioxide gas escapes from this cylinder into a 4.8 L container to bring the pressure of this container to 1.05 atm at 1 C. If the volume of the container then changes, what would the new volume of gas be if its pressure goes to 1.8 atm at 30 C? V 1 = 4.8 L T 1 = 1 ºC + 73.15 = 94 K 1 = 1.05 atm = 1.8 atm T = 30 ºC + 73.15 = 303 K V =? 1 1 T 1 303 K 1.05 atm 1 = 4.8 L = 1.0 L T1 T T1 94 K 1.8 atm
379 CHEMISTRY 151 - IDEAL GAS EQUATION KEY 1. The maximum safe pressure that a certain 4.00 L vessel can hold is 3.50 atm. If the vessel contains 0.410 mole of gas, what is the maximum temperature to which the vessel can be subjected? = nrt ( ) 3.50 atm 4.00 L T = = = 416 K or 143 C nr 0.410 mol 0.08058. A barge containing 640 tons of liquid chlorine had an accident on the Ohio River. What volume would this amount of chlorine occupy if it escaped into the atmosphere as a gas at 756 mmhg and 15 C? g = RT M 640 tons 0.08058 88 K grt 000 lb 453.6 g 760 mmhg V = = 8 = M g 1.95 x 10 L 1 ton 1 lb 1 atm 756 mmhg 70.906 mol 3. What is the density of hydrogen gas at 99.7 K and 386.98 mmhg? g 386.98 mmhg.01594 g g M mol 1 atm = RT = = = M V RT 0.08058 ( 99.7 K) 760 mmhg 0.04174 g/l 4. A radioactive metal atom decays by emitting an alpha particle, which is a helium nucleus. The alpha particles are collected as helium gas. A sample of helium with a volume of 1.05 ml was obtained at 765 mmhg and 3 C. How many atoms decayed during the period of the experiment? = nrt n = RT 3 765 mmhg ( 1.05 ml) 1 atm 1 L 6.0 x 10 atoms? atoms = 3 0.08058 ( 96 K) 760 mmhg 10 ml 1 mol 0 = 3.01 x 10 atoms He
380 Chapter 10 Worksheet Keys GAS STOICHIOMETRY KEY 1. In 1793 the French physicist Jacques Charles supervised and took part in the first human flight in a hydrogen balloon. The hydrogen was produced by reacting iron filings with sulfuric acid to yield hydrogen gas and iron(ii) sulfate. What volume of hydrogen gas at C and 756 mmhg will be formed from the reaction of 500 gallons of 1 M sulfuric acid? Fe(s) + H SO 4 (aq) FeSO 4 (aq) + H (g) 3.785 L 1 mol HSO4 1 mol H 0.08058 4 1 gal 1 L HSO 4 soln 1 mol HSO4? L H = 500 gal H SO 95 K 760 mmhg 5 = 5.5 x 10 L H 756 mmhg 1 atm. Oxygen gas can be generated by heating potassium chlorate to yield oxygen gas and potassium chloride. If 85.0 g of potassium chlorate are reacted in a.50 L container and the oxygen gas is cooled to 1 C in the same container, what will the partial pressure of the oxygen be? KClO 3 (s) KCl(s) + 3O (g)? mol O = 85.0 g KClO 1 mol KClO 3 mol O = 1.04 mol O 3 3 1.553 g KClO3 mol KClO3 1.04 mol O 0.08058 94 K nrt = = = 10.0 atm V.50 L or you can do the calculation in one step. 1 mol KClO 3 mol O 0.08058 94 K? atm = 85.0 g KClO 3 3 = 10.0 atm 1.553 g KClO3 mol KClO3.50 L 3. When carbon disulfide is completely burned, it yields carbon dioxide gas and sulfur dioxide gas. The sulfur dioxide can be removed from the gas mixture by bubbling the mixture through water to dissolve the sulfur dioxide leaving the insoluble carbon dioxide. CS + 3O CO + SO a. What volume of carbon dioxide gas at 19.6 C and 756.4 mmhg will form from the complete combustion of.50 L of carbon disulfide gas at 19.6 C and 756.4 mmhg? 1 L CO? L CO =.50 L CS = 1 L CS.50 L CO b. What volume of carbon dioxide gas at 19.6 C and 756.4 mmhg will form from the complete combustion of.50 L of carbon disulfide gas at 18.1 C and 4.75 atm?? L CO =.50 L CS 4.75 atm 1 mol CO 0.08058 91.3 K 1 mol CS 0.08058 9.8 K 760 mmhg = 1.0 L CO 756.4 mmhg 1 atm
381 c. What volume of carbon dioxide gas at 19.6 C and 756.4 mmhg will form from the complete combustion of.50 L of carbon disulfide gas at 18.1 C and 4.75 atm with 3.5 L of oxygen at 18.1 C and 10. atm?? L CO =.50 L CS 4.75 atm 1 mol CO 0.08058 91.3 K 1 mol CS 0.08058 9.8 K 760 mmhg = 1.0 L CO 756.4 mmhg 1 atm 10. atm 1 mol CO? L CO = 3.75 L O 0.08058 91.3 K 3 mol O 0.08058 9.8 K 760 mmhg = 11. L CO 756.4 mmhg 1 atm This is a limiting reactant problem. The CS would run out when 1.0 L CO are formed. The O runs out when 11. L CO is formed. The O is the limiting reactant.
38 Chapter 10 Worksheet Keys CHEMISTRY 151 - GAS STOICHIOMETRY KEY 1. What volume of dinitrogen oxide gas (nitrous oxide or laughing gas) at 18.0 C and 786.3 mmhg will form from the complete decomposition of 45.987 g ammonium nitrate? NH 4 NO 3 (s) N O(g) + H O(l) 1 mol NH NO 1 mol N O 0.08058 4 3? L NO = 45.987 g NH4NO3 80.0435 g NH 4 NO 3 1 mol NH 4 NO 3 91.15 K 760 mmhg = 13.7 L N O 786.3 mmhg 1 atm. A important reaction in the production of nitrogen fertilizers is the oxidation of ammonia to form nitrogen monoxide and water at 500 C. a. How many liters of nitrogen monoxide will be formed at the same time as 6,500 L of water vapor if both are measured at 500 C and 76.5 mmhg? 4NH 3 + 5O 4NO + 6H O 4 L NO? L NO = 6,500 L HO 6 L HO 4 = 17,667 L NO or 1.77 10 L NO if 6,5000 is three significant figures. b. How many liters of oxygen at 5 C and 0.956 atm are necessary to produce 385 L of NO(g) at 500 C and 765 mmhg? 765 mmhg 1 atm 5 mol O 0.08058 98 K = 195 L O 4 mol NO 0.956 atm? L O = 385 L NO 0.08058 773 K 760 mmhg 3. 34.6 L of carbon dioxide at 34.6 K and 79.4 mmhg form in the complete combustion of naphthalene, C 10 H 8. What volume of oxygen at 75.4 K and 754.9 mmhg must have been consumed in this reaction? C 10 H 8 + 1O 10CO + 4H O 79.4 mmhg 1 atm? L O = 34.6 L CO 0.08058 34.6 K 760 mmhg 1 mol O 0.08058 75.4 K 760 mmhg 10 mol CO 754.9 mmhg 1 atm = 37.0 L O
383 DALTON S LAW OF ARTIAL RESSURES WORKSHEET KEY 1. A sample of hydrogen gas is collected over water. The total pressure of the wet hydrogen in a 500 ml container is 756.5 mmhg at 1 C. If the hydrogen is dried and placed in a 00 ml container at 18 C, what will its pressure be? = + initial = = 756.5 mmhg 18.7 mmhg = 737.8 mmhg total H HO vap H total HOvap 1 1 V1 T 91 K 500 ml = = 1 = 737.8 mmhg 3 = 1.83 x 10 mmhg T1 T V T1 94 K 00 ml To check: decreased V leads to increased, so the volume ratio should be greater than one to lead to the expected increase in pressure. Decreased T leads to decreased, so the temperature ratio should be less than one to lead to the expected decrease in pressure.. A 5 ml container has 0.08 g of hydrogen, 0.15 g of oxygen and 0.09 g of carbon dioxide. The temperature is.5 C. What is the total pressure of the gas in the container? each each RT = = n = n + n + n ( ) total partial H O CO gas gas V RT V 1 mol H 1 mol O 1 mol CO = 0.08 g H + 0.15 g O + 0.09 g CO.0159 g H 31.9988 g O 44.010 g CO 0.08058 ( 95.67 K ) = 4.6 atm 0.5 L 3. A container holds a mixture of 0.0057 g of neon and 0.0081 g of an unknown gas. The partial pressure of the neon is 500.4 mmhg and the total pressure is 755. mmhg. What is the molar mass of the unknown gas? n n n = X = n + n = n = n Ne T Ne T Ne Ne Ne T T Ne unk unk Ne n Ne + nunk Ne Ne 1 mol Ne 755. mmhg 0.0057 g Ne 0.1797 g Ne 1 mol Ne 500.4 mmhg 0.1797 g Ne 4 = - 0.0057 g Ne = 1.4 x 10 mol unk? g unk 0.0081 g unk = = 58 g/mol 4 mol unk 1.4 x 10 mol unk
384 Chapter 10 Worksheet Keys CHEMISTRY 151 - DALTON S LAW OF ARTIAL RESSURES KEY 1. A 500.0 ml sample of a gas is collected over water at 30 ºC and 0.93 atm. What volume would the gas occupy if dry and at 100 ºC and 1.000 atm? 1 atm initial = total water vapor = 0.93 atm 31.84 mmhg = 0.881 atm 760 mmhg 373 K 0.881 atm V = 500.0 ml = 54 ml 303 K 1.000 atm. A cyclopropane-oxygen mixture is used as an anesthetic. Cyclopropane is C 3 H 6. If 4.0 g of cyclopropane and 48.0 g of oxygen are placed in a 5.0 L container at 18.5 ºC, what will the total pressure of the gases be? each RT = n gas V 0.08058 ( 91.7 K 1 mol C ) 3H6 1 mol O = 4.0 g C 3H 6 + 48.0 g O 4.081 g C3H6 31.9988 g O 5.0 L = 9.9 atm 3. A mixture of 0.04 g of N O(g) and 0.85 g NO(g) exerts a pressure of 1.3 atm. What is the partial pressure of each gas? 1 mol NO 0.04 g NO 44.018 g N O X NO = = 0.019 1 mol NO 1 mol NO 0.04 g NO + 0.85 g NO 44.018 g N O 30.0061 g NO = X = 0.019( 1.3 atm ) = 0.05 atm N O NO NO T NO = 1.3 atm 0.05 atm = 1.9 atm NO 4. The atmosphere in a sealed diving bell contained oxygen and helium. If the gas mixture has 0.00 atm of oxygen and a total pressure of 3.00 atm, calculate the mass of helium in 1.00 L of the gas mixture at 0 ºC. = = 3.00 atm 0.00 atm =.80 atm He He T O g.80 atm ( 1.00 L ) 4.006 g M = RT g = = mol = 0.466 g He M RT 0.08058 ( 93 K) 5. A quantity of nitrogen gas originally at 4.60 atm in a 1.0 L container at 3.9 ºC is transferred to a 10.0-L container. A quantity of oxygen originally at 3.50 atm and 3.9 ºC in a 4.00 L container is transferred to the same 10.0 L container with the nitrogen. What is the total pressure of the two gases in the 10.0 L container at 19.7 ºC? total = N + O = 0.544 atm + 1.38 atm = 1.9 atm total 1.0 L 9.9 K N = 4.60 atm = 0.544 atm 10.0 L 97.1 K 4.00 L 9.9 K O = 3.50 atm = 1.38 atm 10.0 L 97.1 K