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Unformatted text preview: nguyen (jn9225) – H03: Mixtures and KMT – mccord – (50950) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Please NOTE: All students in Dr. Mc Cord’s MWF 9am class (50950) will go to our regular classroom of WEL 2.224 for the Exam on Tuesday (79 PM). 001 (part 1 of 4) 10.0 points 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 90 L of oxygen at 3 . 15 atm and 145 ◦ C with an excess of iron pyrite? Correct answer: 239 . 969 g. Explanation: P = 3 . 15 atm T = 145 ◦ C + 273 = 418 K R = 0 . 08206 L · atm K · mol V = 90 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 = nRT n = P V RT = (3 . 15 atm) (90 L) ( . 08206 L · atm K · mol ) (418 K) = 8 . 26505 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 (8 . 26505 mol O 2 ) = 239 . 969 g Fe 2 O 3 . 002 (part 2 of 4) 10.0 points If the sulfur dioxide that is generated above is dissolved to form 6 . 4 L of aqueous solu tion, what is the molar concentration of the resulting sulfurous acid (H 2 SO 3 ) solution? Correct answer: 0 . 93921 M. Explanation: V = 6 . 4 L SO 2 (g) + H 2 O( ℓ )→ H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (8 . 26505 mol) parenleftbigg 8 n SO 2 11 n O 2 parenrightbigg = 6 . 01094 mol . 6 . 01094 mol of SO 2 will dissolve in 6 . 4 L of water to form a solution that is 6 . 01094 mol 6 . 4 L = 0 . 93921 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 highsulfur 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 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 parenrightbiggparenleftbigg 64 . 0638 g SO 2 1 mol SO 2 parenrightbigg = 42718 . 1 g = 42 . 7181 kg SO 2 . nguyen (jn9225) – H03: Mixtures and KMT – mccord – (50950) 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 . 8 atm and 22 ◦ C?...
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This note was uploaded on 09/17/2011 for the course CHEM 301 taught by Professor Wandelt during the Fall '08 term at University of Texas.
 Fall '08
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