90 g What is the molecular formula of the compound 10 points 649 11201 54

# 90 g what is the molecular formula of the compound 10

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compound weighs 6.90 g. What is the molecular formula of the compound? (10 points) 64.9 * 1/12.01 = 5.4 13.5*1/1.008 = 13.4 21.6*1/16= 1.35 Divide all be 1.35 C4H10O Mass= 74.12 225/74.12 = 3 C12H30O3 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 (Reference: Chang 5.59) 73g HCl * 1/36.5 HCl = 2 mol 2 Copyright © 2017 by Thomas Edison State University . All rights reserved. 73g NH3 * 1 / 17 g NH3 = 4.29 mol 2 mol NH4Cl * 53.5 / 1 mol = 107 g of NH4Cl 2.29*.0821*(14+273)/.989 NH3 @ 54.5 L 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) CH4 = .31/.31+.25+.29 = .36 .36*1.5 = .55 C2H6= .25/.25+.31+.29= .29 .29*1.5= .44 C3H8= .29/.29+.31+.29= .34 .34*1.5 =.51 .55+.44+.51 = 1.5 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) C3H8 + 5O2 = 3CO2 + 4H20 Propane equals 7.45 grams = .169 mol .169*3(number of CO2 produced) = .507 1*V= .507*.0821*273 = 11.36 L 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: Moles of Al = 10 * 1/29.982 = .371 Moles of HCl = .54 * .075 = .0405 2:3 Al to H2 = (.371/2) = .185 .185+ .371 = .556 6:3 HCl to H2 = .0405/6 = .00675 .00675* 3 = .0203 3 Copyright © 2017 by Thomas Edison State University . All rights reserved. Volume, in liters, of hydrogen gas. (5 points) .0203*.0821*273*2 .454 L Molarity of Al +3 . (Assume 75.0 mL solution.) (5 points) .0405 HCl = .0135 of AlCl3 .0135/.075= .18 Molarity of Cl . (Assume 75.0 mL solution.) (5 points) .045/.075 = .540 4 Copyright © 2017 by Thomas Edison State University . All rights reserved. #### You've reached the end of your free preview.

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