Homework 05 - Su, Yung Homework 5 Due: Oct 16 2007, 11:00...

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Su, Yung – Homework 5 – Due: Oct 16 2007, 11:00 pm – Inst: FAKHREDDINE 1 This print-out should have 55 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. Each question is worth one point. 001 (part 1 o± 1) 1 points Calculate the number o± ethanol molecules (CH 3 CH 2 OH) that would contain 164 grams o± carbon. 1. 3.73 × 10 25 molec 2. 8.28 × 10 24 molec 3. 4.11 × 10 24 molec correct 4. 5.93 × 10 26 molec 5. 1.89 × 10 26 molec 6. 9.94 × 10 25 molec 7. 1.65 × 10 25 molec Explanation: m C = 164 g ?molec = 164gC × 1molC 12gC × 1 mol C 2 H 6 O 2 mol C × 6 . 022 × 10 23 molec 1molC 2 H 6 O = 4 . 11 × 10 24 molec 002 (part 1 o± 1) 1 points A sample o± pure Fe 2 O 3 contains 14.8 g o± oxygen atoms. Calculate the total grams o± Fe 2 O 3 . The molar mass o± Fe 2 O 3 is 159.69 g/mol. 1. 2.775 g 2. 74.0 g 3. 443 g 4. 0.308 g 5. 12,600 g 6. 49.2 g correct 7. 0.0174 g Explanation: ? g Fe 2 O 3 = 14 . 8 g O × 1 mol O 15 . 9994 g O × 1 mol Fe 2 O 3 3 mol O × 159 . 69 g Fe 2 O 3 1 mol Fe 2 O 3 = 49 . 2 g Fe 2 O 3 003 (part 1 o± 1) 1 points Nitric oxide (NO) is made ±rom the oxidation o± NH 3 according to the equation 4NH 3 + 5O 2 4NO + 6H 2 O . I± an 14 . 87 g sample o± NH 3 gives 16 . 3 g o± NO, what is the percent yield? 1. 51% 2. 100% 3. 98% 4. 62% correct 5. 16% Explanation: m NH 3 = 14 . 87 g m NO = 16 . 3 g The actual yield is 16 . 3 g NO and the theo- retical yield is 14 . 87 g NH 3 × mol NH 3 17 g NH 3 × 4 mol NO 4 mol NH 3 × 30 g NO mol NO = 26 . 2465 g NO Thus the percent yield is ? % Yield = actual yield theoretical yield × 100% = 16 . 3 g NO 26 . 2465 g NO × 100% = 0 . 621034% .
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Su, Yung – Homework 5 – Due: Oct 16 2007, 11:00 pm – Inst: FAKHREDDINE 2 004 (part 1 of 1) 1 points The reaction of 8 . 8 grams of chlorine with ex- cess ±uorine produced 8 grams of ClF 3 . What percent yield of ClF 3 was obtained? Correct answer: 34 . 8591 %. Explanation: m Cl 2 = 8 . 8 g m ClF 3 = 8 g Our ²rst step is to determine the theoretical yield of the reaction. In order to do this, we need to write a balanced chemical equation for the reaction. We know that ±uorine (F 2 ) reacts with chlorine (Cl 2 ) to form ClF 3 : 3 F 2 + Cl 2 2 ClF 3 . The reaction started with 8 . 8 g of Cl 2 . We convert from grams to moles: ? mol Cl 2 = 8 . 8 g Cl 2 × 1 mol Cl 2 70 . 9 g Cl 2 = 0 . 124118 mol Cl 2 . From the balanced equation we see that 1 mol Cl 2 are required to produce 2 mol ClF 3 . We use this mole ratio to calculate the maxi- mum moles of ClF 3 that could be produced: ? mol ClF 3 = 0 . 124118 mol Cl 2 × 2 mol ClF 3 1 mol Cl 2 = 0 . 248237 mol ClF 3 . We convert to grams ClF
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This note was uploaded on 04/15/2009 for the course CH 54740 taught by Professor Fakhreddine during the Spring '09 term at University of Texas at Austin.

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Homework 05 - Su, Yung Homework 5 Due: Oct 16 2007, 11:00...

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