9 g C 135 H 216 g O 100 g of anesthetic C 649g C 12gmol 540 H 135g H 1gmol 135

# 9 g c 135 h 216 g o 100 g of anesthetic c 649g c

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64.9 g C + 13.5 H + 21.6 g O = 100 g of anesthetic C = 64.9g C / 12g/mol = 5.40 H = 13.5g H / 1g/mol = 13.5 O = 21.6g O / 16g/mol = 1.35 Dividing by the smallest one: moles ratio of elements C = 4 H = 10 O = 1 C4H10O1 with molecular weight of 74 224.8 / 74 = 3 then = 3C4H10O 2 Copyright © 2017 by Thomas Edison State University . All rights reserved.
9. What is the mass of the solid NH 4 Cl formed when 73.0 g of NH 3 ( g ) are mixed with an equal mass of gaseous HCl? What is the volume and identity of the gas remaining, measured at 14.0°C and 752 mmHg? (8 points) NH 3 ( g ) + HCl( g ) NH 4 Cl 73g HCl / 36.5HCl =2.0 mol HCl 73g NH3/17g NH3 = 4.29 mol NH3 (Reference: Chang 5.59) 2.0 mol HCl is the limiting reactant 2 mol x 53.5 g NH4Cl = 107g NH4Cl V=Nrt/P with 2.29 moles of NH3 remaining P= 752/760 = 0.989atm V= (2.29 x 0.0821 x 287.15) / 0.989 V= 54.5L NH3 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) Total moles = 0.31+0.25+0.29 = 0.85 CH4= 0.31/0.85 x 1.5 atm = 0.547059 atm 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 moles propane .169 x 3=.508 moles of carbon dioxide 22.44 mol x .508mol= 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: 3 Copyright © 2017 by Thomas Edison State University . All rights reserved.
a. Volume, in liters, of hydrogen gas. (5 points) (0.0405 mol HCl) x (3 mol H2 / 6 mol HCl) x (22.414 L/mol) = 0.45 L H2 b. Molarity of Al +3 . (Assume 75.0 mL solution.) (5 points) (0.0405 mol HCl) x (2 mol Al / 6 mol HCl) / (0.0750 L) = 0.18 mol/L c. Molarity of Cl . (Assume 75.0 mL solution.) (5 points) 0.54 M CI 4 Copyright © 2017 by Thomas Edison State University . All rights reserved.

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