0306 mol n mass m Molar mass M M 69g 0306 mol 2253 gmol 649 2253 gmol 1462 12

0306 mol n mass m molar mass m m 69g 0306 mol 2253

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0306 mol → n = mass (m) / Molar mass (M) → M = 6.9g / .0306 mol = 225.3 g/mol 64.9% * 225.3 g/mol = 146.2 / 12 = 12.2 13.5% * 225.3 g/mol = 30.4 /1 = 30.4 21.6% * 225.3 g/mol = 48.6 /16 = 3 C 12 H 30 O 3 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? NH 3 ( g ) + HCl( g ) NH 4 Cl HCl moles = m / M = 73.0g / 36.5 g/mol = 2 moles NH 3 moles = m / M = 73.0g / 17.034 g/mol = 4.29 moles NH 4 Cl = moles x M = 2 moles * 53.5 g/mol = 107g is the mass of the solid formed If 2 mols of NH 3 reacted, 2.29 moles of it would be left over PV = nRT → V = nRT / P → 2.29 moles * .082 * (14+273)K / 752 mmHg * 1 / 760 mmHg = 54.5 L 2 Copyright © 2018 by Thomas Edison State University. All rights reserved. 1.2 atm 2.5 L 250 K 298 K 3e-3 atm 4.65 g .082 L atm 300 K 1 atm K mol 2.1 L
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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. .31 + .25 + .29 = .85 moles .31 / .85 = .365 * 1.5 atm = .546 atm .25 / .85 = .294 * 1.5 atm = .441 atm .29 / .85 = .341 * 1.5 atm = .512 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. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O C 3 H 8 moles = mass C 3 H 8 / M C 3 H 8 = 7.45g / 44.09 = .169 moles 3 * .169 moles = moles of CO 2 = .507 moles At STP 1 mole of gas has a volume of 22.4 L = 22.4 L * .057 moles = 11.36 L of CO 2 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 Moles of Al = m / M = 10 g / 26.98 g/mol = .371 moles Moles of HCl = molarity * V = .54 moles * 75 mL = .0405 moles HCl .371 / 2 = .1852 moles Al .0405 / 6 = .00675 moles HCl → 6 moles HCl will produce 3 mole of H 2 = .02025 mol Hydrogen = .45 L of H 2 .0405 * 1/3 / .075 L = .18 M a. Volume, in liters, of hydrogen gas. .45 L b. Molarity of Al +3 . (Assume 75.0 mL solution.) .18 M c. Molarity of Cl . (Assume 75.0 mL solution.) .54 M because the Cl concentration remained the same throughout the reaction 3 Copyright © 2018 by Thomas Edison State University. All rights reserved. 6 mol HCl .0405 mol HCl 3 mol H X mol H 1 mol H 22.4 L .02025 X L
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