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Basic mole calculations (pp.112-124) Exercise 3

Where we’ve been for ~ 3 weeks now:. Basic mole calculations (pp.112-124) Exercise 3 % composition problems/combustion analysis (pp.384-392) Reaction balancing (pp. 392-395) Reaction stoichiometry predictions (pp. 396-406 ) Exercise 4. The final assault on mole calculations .

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Basic mole calculations (pp.112-124) Exercise 3

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  1. Where we’ve been for ~ 3 weeks now: • Basic mole calculations (pp.112-124) Exercise 3 • % composition problems/combustion analysis (pp.384-392) • Reaction balancing (pp. 392-395) • Reaction stoichiometry predictions (pp. 396-406) Exercise 4 The final assault on mole calculations Limiting yield and % yield calculations pp. 400-406

  2. A non-chemical example of a `limiting’ yield problem • You have gotten lost on the Ad Dahna desert –largest desert on the Saudi peninsula. To avoid perishing you must reach the nearest oasis which is 100 km away. You can walk at maximum 10 km/ day. You need to consume at least 2 liters of water and ½ kg of food per day to walk that distance. • You have in your pack: • 16 liters of water • 5 kg of food • Broken cell phone What’s the maximum distance you can expect to travel ??

  3. Food Calculation 5 kg food = 10 days ½ kg/day =>10 days * 10 km = 100 km  day

  4. Water Calculation 16 liters = 8 days 2 liters/day • 8 days * 10 km = 80 km  • day The winner is….always the smaller one…it limits.

  5. Examples of chemical limiting yield calculations Chemical reaction example #1 5O2 + C3H83CO2 + 4H2O mol 0.33 0.05 ?? mol Given 0.33 mol O2 and 0.10 mol C3H8, compute the maximum theoretical yield of CO2moles Ans. 0.15 mol CO2(C3H8 limits)

  6. The `cut & try’ approach to limiting yield calculations (cont.) Chemical reaction example #2: 0.18 g H2O (O2 limits) 5O2 + C3H83CO2 + 4H2O g 4.0 2.2 ? g MW 32 4444 18 Given 4.0 grams O2 and 2.2 grams C3H8, compute the theoretical yield of H2O for the combustion shown above

  7. C12H22O11 + 12O2 -------- 12CO2 + 11H2O MW 342 g/mol 32 g/mol 44 g/mol 18 g/mol w(g)68.40 72.73 ?? g • How many grams of CO2 will be produced if 68.40 grams of sucrose, C12H22O11, is • combined with 72.73 grams of O2 according to the balanced equation above? 100 g CO2 (O2 limits)

  8. EXERCISE 5 Practice Examples first done in class-on white board

  9. 5. 1. ID the limiting reagent • 2C4H10 + 13O2-------- 8CO2 + 10H2O • 58 32 44 18 g/mol • a) 1 mole of C4H10and 6 moles of O2are reacted. Limiting = C4H10 or O2? 1) Pick H2O to predict moles 2) Predict moles produced from both C4H10and O2 moles Moles H2O = 10 = m(H2O) Moles C4H102 1 m(H2O) = 5 Moles H2O = 10 = m(H2O) Moles O213 6 m(H2O) = 60/13=4.6 O2 Runs out first

  10. 5. 1. ID the limiting reagent (cont.) • 2C4H10 + 13O2-------- 8CO2 + 10H2O • 58 32 44 18 g/mol • b) 10 grams of C4H10 and 100 grams of O2 are reacted. Limiting = C4H10 or O2 ? 1) Convert mass moles (divide up) Mol C4H10 = 10/58=0.17 Mol O2 = 100/32=3.125 2) Predict moles produced from both C4H10and O2 moles Moles H2O = 10 = m(H2O) Moles C4H102 0.17 m(H2O) = 10*0.17 =0.425 2 C4H10 Runs out first Moles H2O = 10 = m(H2O) Moles O213 3.125 m(H2O) = 10*3.125 =2.4 13

  11. 5. 1. ID the limiting reagent (cont.) • 2C4H10 + 13O2-------- 8CO2 + 10H2O • 58 32 44 18 g/mol • c) 2*1022 molecules of C4H10 and 50 grams of O2 are reacted. Limiting = C4H10 or O2 ? 1) Convert mass, count moles (divide up) Mol C4H10 = 2*1022/6*1023=0.033 Mol O2 = 50/32=1.56 2) Predict moles produced from both C4H10and O2 moles Moles H2O = 10 = m(H2O) Moles C4H102 0.033 m(H2O) = 10*0.033 =0.15 2 C4H10 runs out first Moles H2O = 10 = m(H2O) Moles O213 1.56 m(H2O) = 10*1.56 =1.2 13

  12. 5.2 Calculate the maximum yield problems 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Mole-mol-mol • a)0.25 moles of C4H10and 1.4 moles of O2 are reacted. • What is the maximum yield of CO2 in moles? 0.25 mol C4H10 produces 8*0.25/2 =1 mol CO2 1.4 mol O2 produces 8*0.25/13 =0.15 mol CO2

  13. 5.2 Calculate the maximum yield problems (cont.) 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol • Weight-weight-mol • b)1 gram of C4H10 and 10 grams of O2 are reacted. • What is the maximum yield in H2O in moles ? 1 gram C4H10= 1/58=0.017 mol C4H10 10 gram O2= 10/32=0.312 mol O2 0.017 mol C4H10 produces 10*0.017/2=0.085 mol H2O 0.312 mol O2 produces 10*0.312/13 =0.24 mol H2O

  14. 5.2 Calculate the maximum yield problems (cont.) 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Weight-weight-molecules • c) 5.8 grams of C4H10and 160 grams of O2 are reacted. • What is the maximum yield of CO2in molecule count? 5.8/58=0.1 mol C4H10 160/32=5 mol O2 0.1 mol C4H10 produces 8*0.1/2=0.4 mol CO2 5 mol O2 produces 8*5/13 =3.07 mol CO2 C4H10limits 0.4*6*1023 = molecules CO2= 2.4*1023

  15. 5.2 Calculate the maximum yield problems (cont.) 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Weight-molecules-weight • d)116 grams of C4H10 and 1.66*1024 molecules of O2 react. • What is the maximum yield of H2O in grams? 116/58=2 mol C4H10 1.66*1024/6*1023=1.1 mol O2 2 mol C4H10 produces 10*2/2=10 mol H2O 1.1 mol O2 produces 10*1.1/13 =0.85 mol H2O mol O2limits Multiply down to calculate: grams H2O= 0.85*18=15.3 g

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