Chem Lab Quizzes and Post-Labs

Chem Lab Quizzes and Post-Labs - goins (jmg3964) Post-lab 4...

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Unformatted text preview: goins (jmg3964) Post-lab 4 lyon (51215) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 33 . 6 g sample of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to Fe 2 O 3 . The mass of Fe 2 O 3 is measured to be 8 . 1 g. What was the mass of iron in the sample of ore? Correct answer: 5 . 66545 g. Explanation: m iron ore = 33 . 6 g m Fe 2 O 3 = 8 . 1 g 8 . 1 g Fe 2 O 3 1 mol Fe 2 O 3 159 . 691 g Fe 2 O 3 2 mol Fe 1 mol Fe 2 O 3 55 . 847 g Fe 1 mol Fe = 5 . 66545 g Fe 002 10.0 points Determine the limiting reactant for the fol- lowing reaction, given that 100 grams of each reactant was used. 3 Cu(s) + 2 H 2 O(l) + SO 2 (g) + 2 O 2 (g) Cu 3 (OH) 4 SO 4 (s) 1. SO 2 (g) - 0.525 moles of product formed 2. Cu (s) - 1.56 moles of product formed 3. H 2 O (l) - 2.77 moles of product formed 4. O 2 (g) - 2.77 moles of product formed 5. H 2 O (l) - 0.525 moles of product formed 6. SO 2 (g) - 1.56 moles of product formed 7. Cu (s) - 0.525 moles of product formed correct 8. O 2 (g) - 1.56 moles of product formed Explanation: There are many different ways to do lim- iting reagent problems. This is just one way you could do it. 1) 100 grams of copper metal divided by its atomic weight (63.546 g/mole) gives you moles of copper. According to the balanced equation, you will get 1 mole of Cu 3 (OH) 4 SO 4 for every 3 moles of Cu, so the moles of Cu 3 (OH) 4 SO 4 formed is equal to 1/3 the moles of Cu. If you do the math, 100 g of Cu is enough to form 0.525 moles of product. 2) 100 grams of water divided by its molec- ular weight (18.0152 g/mole) gives you moles of water. According to the balanced equation, you will get 1 mole of Cu 3 (OH) 4 SO 4 for every 2 moles of H 2 O, so the moles of Cu 3 (OH) 4 SO 4 formed is equal to 1/2 the moles of H 2 O. 100 g of H 2 O is enough to form 2.77 moles of product. 3) 100 grams of SO 2 divided by its molec- ular weight (64.0588 g/mole) gives you moles of sulfur dioxide. According to the balanced equation, you will get 1 mole of Cu 3 (OH) 4 SO 4 for every mole of SO 2 , so the moles of Cu 3 (OH) 4 SO 4 formed is equal to the num- ber of moles of SO 2 . 100 g of SO 2 is enough to form 1.56 moles of product. 4) Do the same thing with O 2 and youll find that 100 g of O 2 will also form 1.56 moles of product. Whichever one of these four reagents gives you the fewest moles of product is the limiting reactant. In this case the copper metal runs out first, after forming 0.525 moles of product. 003 10.0 points Which of these best describes the concept of a limiting reagent? 1. The limiting reagent is the reactant that determines the rate of a reaction....
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Chem Lab Quizzes and Post-Labs - goins (jmg3964) Post-lab 4...

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