1 mol C H OH 3 mol O 227 g C H OH 148 mol O 4607 g C H OH 1 mol C H OH 148

1 mol c h oh 3 mol o 227 g c h oh 148 mol o 4607 g c

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1 mol C H OH 3 mol O 227 g C H OH 14.8 mol O 46.07 g C H OH 1 mol C H OH 14.8 moles of O 2 correspond to a volume of: 2 2 2 O 2 O 2 L atm (14.8 mol O ) 0.0821 (35 273 K) mol K 3.60 10 L O 1 atm 790 mmHg 760 mmHg n RT V P Since air is 21.0 percent O 2 by volume, we can write: 2 2 O 2 2 2 100% air 100% air (3.60 10 L O ) 21% O 21% O 3 air 1.71 10 L air V V 5.63 (a) 3 4 L NH 12.8 L NO 4 L NO 3 12.8 L NH 2 5 L O 12.8 L NO 4 L NO 2 16.0 L O (b) 3 8 2 2 1 L C H 8.96 L H 7 L H 3 8 1.28 L C H 2 2 2 3 L H O 8.96 L H 7 L H 2 3.84 L H O 5.64 The balanced equation is: FeS + 2HCl H 2 S + FeCl 2 . From the moles of H 2 S produced, we can calculate the mass of FeS in the 4.00 g sample. The mass percent purity can then be calculated. 2 H S 2 1 atm 782 mmHg (0.896 L) 760 mmHg 0.0391 mol H S L atm 0.0821 (287 K) mol K PV n RT The mass of FeS in the 4.00 g sample is: 2 2 1 mol FeS 87.92 g FeS 0.0391 mol H S 3.44 g FeS 1 mol H S 1 mol FeS
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CHAPTER 5: GASES 150 The mass percent purity is: 3.44 g 100% 4.00 g mass percent purity 86.0% 5.67 First, we calculate the mole fraction of each component of the mixture. Then, we can calculate the partial pressure of each component using the equation, P i i P T . The number of moles of the combined gases is: 4 2 6 3 8 CH C H C H 0.31 mol 0.25 mol 0.29 mol 0.85 mol n n n n 4 2 6 3 8 CH C H C H 0.31 mol 0.25 mol 0.29 mol 0.36 0.29 0.34 0.85 mol 0.85 mol 0.85 mol The partial pressures are: 4 CH total 0.36 1.50 atm 4 CH 0.54 atm P P 2 6 C H total 0.29 1.50 atm 2 6 C H 0.44 atm P P 3 8 C H total 0.34 1.50 atm 3 8 C H 0.51 atm P P 5.68 Dalton's law states that the total pressure of the mixture is the sum of the partial pressures. (a) P total 0.32 atm 0.15 atm 0.42 atm 0.89 atm (b) We know: Initial conditions Final Conditions P 1 (0.15 0.42)atm 0.57 atm P 2 1.0 atm T 1 (15 273)K 288 K T 2 273 K V 1 2.5 L V 2 ? 1 1 2 2 1 1 2 2 PV P V n T n T Because n 1 n 2 , we can write: 1 1 2 2 2 1 PV T V P T (0.57 atm)(2.5 L)(273 K) (1.0 atm)(288 K) 2 1.4 L at STP V 5.69 Since volume is proportional to the number of moles of gas present, we can directly convert the volume percents to mole fractions. 2 2 2 N O Ar CO 0.7808 0.2094 0.0093 0.0005 (a) For each gas, P i i P T i (1.00 atm). , , , 2 2 2 -3 -4 N O Ar CO 0.781 atm 0.209 atm 9.3 × 10 atm 5 × 10 atm P P P P
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CHAPTER 5: GASES 151 (b) Concentration (mol/L) is n P c V RT . Therefore, we have: 0.781 atm L atm 0.0821 (273 K) mol K 2 2 N 3.48 10 c M Similarly, , , 2 2 3 4 5 O Ar CO 9.32 10 4.1 10 2 10 c M c M c M 5.70 P Total P 1 P 2 P 3 . . . P n In this case, 2 Total Ne He H O + + P P P P 2 Ne Total He H O P P P P P Ne 745 mm Hg 368 mmHg 28.3 mmHg 349 mmHg 5.71 If we can calculate the moles of H 2 gas collected, we can determine the amount of Na that must have reacted.
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