moles massmolecular weight 00307 moles 690molecular weight Molecular weight

Moles massmolecular weight 00307 moles 690molecular

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moles= mass/molecular weight 0.0307 moles= 6.90/molecular weight Molecular weight= 224.8 moles C: 64.9 g C/12 g/mole C=5.40 moles H: 13.5 g H/1g/mole H= 13.5 moles O: 21.6 g O/16 g/mole O= 1.35 moles ratio of elements: C 5.40/1.35=4; H 13.5/1.35=10; O 1.35/1.35= 1 C4H10O1 and molecular weight = 74 224.8/74=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 (73.0 g HCl)(1 mol HCl / 36.5g HCl = 2.00 mol (73.0 g NH3)(1 mol NH3 / 17.0 g HCl) = 4.29 mol 2.00 moles of NH4Cl will be produced. (2.00 mol NH4Cl)(53.5 g NH4Cl) / 1 mol NH4Cl = 107.0 g NH4Cl PV = nRT V = nRT / P V =((2.29 mol)(62.4 L mmHg) / mol K))( (14+273 K) / 752 mm Hg) = 54.5 L 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) total moles of gas = 0.31+0.25+0.29 = 0.85 partial pressure of CH4 = (0.31/0.85)(1.5 atm) = 0.547 atm partial pressure of C2H6 = (0.25/0.85)(1.5 atm) = 0.441 atm partial pressure of C3H8 = (0.29/0.85)(1.5 atm) = 0.512 atm 3 Copyright © 2014 by Thomas Edison State College. All rights reserved.
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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) A. C3H8g+5O2g = 3CO2g+4H2Og B. CO2 L= (7.45)(.02268)(3)(22.14L) =11.4 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: a. Volume, in liters, of hydrogen gas. (5 points) (0.0405 mol HCl)(3 mol H2) / (6 mol HCl)(22.414 L/mol) = 0.45 L H2 b. Molarity of Al +3 . (Assume 75.0 mL solution.) (5 points) ((0.0405 mol HCl)(2 mol Al3) / 6 mol HCl) / (0.0750 L) = 0.18 mol/L Al3 c. Molarity of Cl . (Assume 75.0 mL solution.) (5 points) The concentration of the Cl ions is the same as the original concentration of the HCl since the volume didn't change, so 0.54 M Cl 4 Copyright © 2014 by Thomas Edison State College. All rights reserved.
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