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HW 03 Mixtures and Kinetic Theory-solutions

HW 03 Mixtures and Kinetic Theory-solutions - nanni(arn437...

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nanni (arn437) – HW 03 Mixtures and Kinetic Theory – vanden bout – (50985) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 4) 10.0 points Iron pyrite (±eS 2 ) is the Form in which much oF the sulFur exists in coal. In the combustion oF coal, oxygen reacts with iron pyrite to produce iron(III) oxide and sulFur dioxide, which is a major source oF air pollution and a substantial contributor to acid rain. What mass oF ±e 2 O 3 is produced From the reaction is 62 L oF oxygen at 3 . 27 atm and 165 C with an excess oF iron pyrite? Correct answer: 163 . 774 g. Explanation: P = 3 . 27 atm T = 165 C + 273 = 438 K R = 0 . 08206 L · atm K · mol V = 62 L MW Fe 2 O 3 = 2(55 . 845 g / mol) + 3(15 . 9994 g / mol) = 159 . 688 g / mol The balanced equation is 4 ±eS 2 (s) + 11 O 2 (g) -→ 2 ±e 2 O 3 (s) + 8 SO 2 (g) Applying the ideal gas law to the O 2 , P V = nRT n = P V RT = (3 . 27 atm) (62 L) ( 0 . 08206 L · atm K · mol ) (438 K) = 5 . 64071 mol . ±rom stoichiometry and the molar mass oF ±e 2 O 3 , m 2 O 3 = (159 . 688 g / mol ±e 2 O 3 ) × 2 mol ±e 2 O 3 11 mol O 2 (5 . 64071 mol O 2 ) = 163 . 774 g ±e 2 O 3 . 002 (part 2 of 4) 10.0 points IF the sulFur dioxide that is generated above is dissolved to Form 6 . 6 L oF aqueous solu- tion, what is the molar concentration oF the resulting sulFurous acid (H 2 SO 3 ) solution? Correct answer: 0 . 621566 M. Explanation: V = 6 . 6 L SO 2 (g) + H 2 O( ) H 2 SO 3 (aq) . ±rom the stoichiometry, n SO 2 = (5 . 64071 mol) p 8 n SO 2 11 n O 2 P = 4 . 10233 mol . 4 . 10233 mol oF SO 2 will dissolve in 6 . 6 L oF water to Form a solution that is 4 . 10233 mol 6 . 6 L = 0 . 621566 M in H 2 SO 4 . 003 (part 3 of 4) 10.0 points What mass oF SO 2 is produced in the burning oF 1 tonne (1 t = 1000 kg) oF high-sulFur coal, iF the coal is 4% pyrite by mass? Correct answer: 42 . 7181 kg. Explanation: m coal = 1000 kg m FeS 2 = 4%(1000 kg) = 40 kg = 40000 g MW 2 = 55 . 845 g / mol + 2(32 . 065 g / mol) = 119 . 975 g MW SO 2 = 32 . 065 g / mol + 2(15 . 9994 g / mol) = 64 . 0638 g m SO 2 = 1000 kg coal p 40000 g ±eS 2 1000 kg coal P × p 1 mol ±eS 2 119 . 975 g ±eS 2 P × p 8 mol SO 2 4 mol ±eS 2 Pp 64 . 0638 g SO 2 1 mol SO 2 P = 42718 . 1 g = 42 . 7181 kg SO 2 .
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nanni (arn437) – HW 03 Mixtures and Kinetic Theory – vanden bout – (50985) 2 004 (part 4 of 4) 10.0 points From the previous problem, what is the vol- ume of the SO 2 gas at 1 atm and 44 C? Correct answer: 17345 . 6 L. Explanation: R = 0 . 08206 L · atm K · mol T = 44 C + 273 = 317 K P = 1 atm The number of moles of SO 2 is n SO 4 = (42718 . 1 g SO 2 ) 1 mol SO 2 64 . 0638 g SO 2 = 666 . 806 mol SO 2 . From the ideal gas law, V = nT R P = (666 . 806 mol) (317 K) 1 atm × p 0 . 08206 L · atm K · mol P = 17345 . 6 L . 005 10.0 points Two gases are contained in gas bulbs con- nected by a valve. Gas A is present in a 1 liter bulb at a pressure of 835 torr. Gas B ex- erts a pressure of 202 torr in a 1 liter bulb. The valve is opened and the two gases equili- brate. What is the partial pressure of gas A expressed after equilibration? Correct answer: 417 . 5 torr. Explanation: V A = 1 L V B = 1 L P A = 835 torr P B = 202 torr V total = 2 L P A V A = P t V t P t = P A V A V t = (835 torr) (1 L) 2 L = 417 . 5 torr 006 10.0 points A reminder. .. partial pressures are calculated from the total pressure via mole fractions, not percent by mass. So you WILL have to convert the percent by mass to mole fraction BEFORE you answer this problem.
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