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# hw3ans - wilson(tw8656 H03 Mixtures and KMT mccord(50960...

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wilson (tw8656) – H03: Mixtures and KMT – mccord – (50960) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. All students from Dr. McCord’s MWF 11am class (50960) must report to the correct room for the exams. There are 2 different rooms and the class is split according to the first letter of your last name. Go to the right room or you WILL receive a zero on the exams. A-L go to UTC 2.112A M-Z go to UTC 2.102A 001(part1of4)10.0points Iron pyrite (FeS 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 Fe 2 O 3 is produced from the reaction is 77 L of oxygen at 3 . 58 atm and 149 C with an excess of iron pyrite? Correct answer: 231 . 121 g. Explanation: P = 3 . 58 atm T = 149 C + 273 = 422 K R = 0 . 08206 L · atm K · mol V = 77 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 FeS 2 (s) + 11 O 2 (g) -→ 2 Fe 2 O 3 (s) + 8 SO 2 (g) Applying the ideal gas law to the O 2 , P V = n R T n = P V R T = (3 . 58 atm) (77 L) ( 0 . 08206 L · atm K · mol ) (422 K) = 7 . 96031 mol . From stoichiometry and the molar mass of Fe 2 O 3 , m Fe 2 O 3 = (159 . 688 g / mol Fe 2 O 3 ) × 2 mol Fe 2 O 3 11 mol O 2 (7 . 96031 mol O 2 ) = 231 . 121 g Fe 2 O 3 . 002(part2of4)10.0points If the sulfur dioxide that is generated above is dissolved to form 5 . 5 L of aqueous solu- tion, what is the molar concentration of the resulting sulfurous acid (H 2 SO 3 ) solution? Correct answer: 1 . 0526 M. Explanation: V = 5 . 5 L SO 2 (g) + H 2 O( ) -→ H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (7 . 96031 mol) parenleftbigg 8 n SO 2 11 n O 2 parenrightbigg = 5 . 78931 mol . 5 . 78931 mol of SO 2 will dissolve in 5 . 5 L of water to form a solution that is 5 . 78931 mol 5 . 5 L = 1 . 0526 M in H 2 SO 4 . 003(part3of4)10.0points 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 FeS 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

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wilson (tw8656) – H03: Mixtures and KMT – mccord – (50960) 2 m SO 2 = 1000 kg coal parenleftbigg 40000 g FeS 2 1000 kg coal parenrightbigg × parenleftbigg 1 mol FeS 2 119 . 975 g FeS 2 parenrightbigg × parenleftbigg 8 mol SO 2 4 mol FeS 2 parenrightbigg parenleftbigg 64 . 0638 g SO 2 1 mol SO 2 parenrightbigg = 42718 . 1 g = 42 . 7181 kg SO 2 . 004(part4of4)10.0points From the previous problem, what is the vol- ume of the SO 2 gas at 0 . 8 atm and 34 C? Correct answer: 20998 . 1 L. Explanation: R = 0 . 08206 L · atm K · mol T = 34 C + 273 = 307 K P = 0 . 8 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 = n T R P = (666 . 806 mol) (307 K) 0 . 8 atm × parenleftbigg 0 . 08206 L · atm K · mol parenrightbigg = 20998 . 1 L .
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