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hmwk 3 solutions

hmwk 3 solutions - mao(tm23477 Homework 3 Ideal gases...

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mao (tm23477) – Homework 3: Ideal gases – Shear – (52375) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. You may need to look up conversion Factors and values oF R in the back oF your textbook. 001 (part 1 oF 3) 10.0 points Methanol (CH 3 OH) is the simplest oF the al- cohols. It is synthesized by the reaction oF hydrogen and carbon monoxide: CO(g) + 2 H 2 (g) -→ CH 3 OH IF 465 mol oF CO and 797 mol oF H 2 are present, which is the limiting reactant? 1. CO 2. H 2 correct Explanation: n CO = 465 mol n H 2 = 797 mol limiting reactant = ? mole ratio : 2 mol H 2 1 mol CO . n H 2 required = 465 molCO × 2 mol H 2 1 mol CO = 930 mol H 2 Because there is only 797 mol H 2 available, the limiting reactant is H 2 . 002 (part 2 oF 3) 10.0 points How much oF the excess reactant remains un- changed? Correct answer: 66 . 5 mol. Explanation: remaining excess reactant = ? n CO reacting = 1 mol CO 2 mol H 2 × 797 mol H 2 = 398 . 5 mol CO n excessreactant = 465 mol CO - 398 . 5 mol CO = 66 . 5 mol CO 003 (part 3 oF 3) 10.0 points How much CH 3 OH is Formed? Correct answer: 398 . 5 mol. Explanation: n limiting reactant = 797 mol H 2 amounts oF CH 3 OH = ? mole ratio : 1 mol CH 3 OH 2 mol H 2 n CH 3 OH = 1 mol CH 3 OH 2 mol H 2 × 797 mol H 2 = 398 . 5 mol CH 3 OH 004 10.0 points Helium is used to in±ate Football-shaped bal- loons that will be released in Memorial sta- dium For the opening game oF the Football season. One 2.00 liter balloon pressurized

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hmwk 3 solutions - mao(tm23477 Homework 3 Ideal gases...

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