987 atm x 100 L0082 x 3923 K 00307 mols mols massmw 00307 mols 690 mw mw 2248

# 987 atm x 100 l0082 x 3923 k 00307 mols mols massmw

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n=PV/RT= (0.987 atm x 1.00 L)/(0.082 x 392.3 K) = 0.0307 mols mols = mass/mw 0.0307 mols = 6.90 / mw mw = 224.8 100 g anesthetic: 64.9 g C + 13.5 H + 21.6 g O mols C: 64.9 g C/12 g/mol C=5.40 mols H: 13.5 g H/1g/mol H= 13.5 mols O: 21.6 g O/16 g/mol O= 1.35 mols ratio of elements: C 5.40/1.35=4; H 13.5/1.35=10; O 1.35/1.35= 1 C4H10O1 mw = 74 224.8/74=~3; C12H30O3 9.What is the mass of the solid NH4Cl formed when 73.0 g of NH3(g) are mixed with an equal mass of gaseous HCl? What is the volume and identity of the gas 2
10. A mixture of gases contains 0.31 mol CH 4 , 0.25 mol C 2 H 6, and 0.29 mol C 3 H 8 . The total pressure is 1.50 atm. Calculate the partial pressures of the gases. (8 points) (Reference: Chang 5.67) mols = 0.31 + 0.25 + 0.29 = 0.85 CH4 = 0.31 / 0.85 x 1.5 atm = 0.547059 atm sig. fig. errors -3 C2H6 = 0.25 / 0.85 x 1.5 atm = 0.441176 atm C3H8 = 0.29 / 0.85 x 1.5 atm = 0.511765 atm 11. Propane (C 3 H 8 ) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from 7.45 g of propane. (10 points) (Reference: Chang 5.109) C3H8+5O2 ---- 3CO2 + 4H20 7.45g x 1mol / 44g/mol = .169 mols propane 0.169 x 3 =0.508 mols of CO2 22.44 mol x 0.508 mol = 11.4L 12. A 10.0 g piece of pure aluminum is placed in 75.0 mL of 0.54 M hydrochloric acid at STP condition. They react as follows: 2Al + 6HCl 3H 2 ( g ) + 2AlCl 3 Calculate the following: a. Volume, in liters, of hydrogen gas. (5 points) b. Molarity of Al +3 . (Assume 75.0 mL solution.) (5 points) c. Molarity of Cl . (Assume 75.0 mL solution.) (5 points) a. (0.0405 mol HCl) x (3 mol H2 / 6 mol HCl) x (22.414 L/mol) = 0.45 L H 2 b. (0.0405 mol HCl) x (2 mol Al / 6 mol HCl) / (0.0750 L) = 0.18 mol/L 3 Copyright © 2016 by Thomas Edison State University . All rights reserved.

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