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c) weight to moles How many moles of H 2 O form if 11 g of C 3 H 8 are burned ?

C 3 H 8 + 5O 2 --------  3CO 2 + 4H 2 O (BOOM) 44 32 44 18 g/mol . c) weight to moles How many moles of H 2 O form if 11 g of C 3 H 8 are burned ? 1) convert non-moles to moles 11 g C 3 H 8 * 1 mol C 3 H 8 44 g C 3 H 8

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c) weight to moles How many moles of H 2 O form if 11 g of C 3 H 8 are burned ?

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  1. C3H8 + 5O2--------3CO2 + 4H2O (BOOM) • 44 32 44 18 g/mol • c) weight to moles • How many moles of H2O form if 11 g of C3H8 are burned ? • 1) convert non-moles to moles • 11 g C3H8* 1 mol C3H8 • 44 g C3H8 • = 0.25 mol C3H8 • 2) set up mole ratios • mol H2O = 4 = m(H2O) • mol C3H8 1 0.25 • Solve for m • 4*0.25 = m(H2O) =1 mol H2O • 5

  2. C3H8 + 5O2--------3CO2 + 4H2O (BOOM) • 44 32 44 18 g/mol d) weight to weight How many grams of O2are needed to burn 0.275 g C3H8 ? 0.275/44=0.00625 mol C3H8 1) Convert gmole 2a)Ratio target mol/known mol Mol O2/mol C3H8=5/1 = m(O2)/0.00625 2b) solve for target moles, m m(O2) = 5*0.00625 = 0.03125 3) Convert target moles to mass desired Weight O2 = m(O2)*MW(g/mol) =0.03125*32 = 1 g O2

  3. C3H8 + 5O2--------3CO2 + 4H2O (BOOM) • 44 32 44 18 g/mol • e) weight to count • How many molecules of CO2form if 0.398 g H2O results ? • 0.398 g H2O *1 mol H2O = 0.02211 mol H2O • 18 g H2O • m(CO2) = 3 = m(CO2) • m(H2O) 4 0.02211 m(CO2)= 0.0165 m(CO2)= 3*0.02211 4 =1*1022 Molecule count = m(CO2) *6.02*1023

  4. C3H8 + 5O2--------3CO2 + 4H2O (BOOM) • 44 32 44 18 g/mol f) count to weight How many grams of O2are needed to form 1.50*1022 molecules of H2O ? 1.5*1022 6.02*1023 Mole H2O = =0.02491 mol H2O Mol O2 Mol H2O = 5 = m(O2) 4 0.02491 m(O2)= 5*0.02491 4 =0.03114 Weight O2= m(O2)*MW (g/mol) = 1 g O2 =0.03114*32

  5. Sample Reaction 2 6HCl + 2Al ----- 2AlCl3 + 3H2 36 27 123 2 g/mol moles to moles: How many moles of Al must be added to produce 15 moles of H2 ? 10 mol Al b)moles to weight: How many grams of H2are created by reacting 10 moles of HCl ? 10 grams H2

  6. Sample Reaction 2 (home practice) 6HCl + 2Al ----- 2AlCl3 + 3H2 36 27 123 2 g/mol c) weight to moles: How many moles of HClcan combine with 90 g of Al ? 10 moles HCl d) weight to weight: How many grams of Al must react to form 1.1111 grams of H2 ? 10 grams Al

  7. MW 58 32 44 18Given: 2C4H10 + 13O28CO2 +10H2OHow many grams of H2O result from burning 12.888 g C4H10 ? • 0.222 g • 1.111 g • 10. 0 g • 20.0 g • 39.1 g • None of above

  8. MW=16 32 44 18Given CH4 + 2O2 CO2 + 2H2O (1 mol count=6*1023)How many molecules of CO2form when 53.33 g O2 are burned in the above reaction ? • 2.5*1023 • 5*1024 • 1*1024 • 5*1023 • NONE OF ABOVE

  9. Where we’ve been for ~ 2 weeks now: • Basic mole calculations (pp.112-124) • % composition problems/combustion analysis (pp.384-392) • Reaction balancing (pp. 392-395) • Reaction stoichiometry predictions (pp. 396-406) The final assault on mole calculations Limiting yield and % yield calculations pp. 400-406

  10. A non-chemical example of a `limiting’ yield problem • You have gotten lost on the Ad Dahna desert –largest desert on the Saudi pennisula. 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 ??

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

  12. 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.

  13. 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)

  14. 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

  15. 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)

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