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9. 2 C 3 H 7 OH + 9 O 2 → 6 CO 2 + 8 H 2 O

9. 2 C 3 H 7 OH + 9 O 2 → 6 CO 2 + 8 H 2 O. Identify some mole ratios, using the given equation for the combustion of isopropyl alchol. a) 6 mol CO 2 2 mol C 3 H 7 OH. d) 8 mol H 2 O 2 mol C 3 H 7 OH. b ) 2 mol C 3 H 7 OH 9 mol O 2. e ) 6 mol CO 2

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9. 2 C 3 H 7 OH + 9 O 2 → 6 CO 2 + 8 H 2 O

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  1. 9. 2 C3H7OH + 9 O2 → 6 CO2 + 8 H2O Identify some mole ratios, using the given equation for the combustion of isopropyl alchol. a) 6 mol CO2 2 mol C3H7OH d) 8 mol H2O 2 mol C3H7OH b) 2 mol C3H7OH 9 mol O2 e) 6 mol CO2 8 mol H2O c) 9 mol O2 6 mol CO2 f) 8 mol H2O 9 mol O2

  2. Mole Ratios 3 CaCl2 + 2 H3PO4 → Ca3(PO4)2 + 6 HCl Identify some mole ratios, using the given equation. a) 3 CaCl2 b) 6 HCl c) 3 CaCl2 1 Ca3(PO4)2 2 H3PO4 2 H3PO4 d) 1 Ca3(PO4)2 e) 6 HCl f) 2 H3PO4 2 H3PO4 1 Ca3(PO4)2 6 HCl

  3. 11. C2H5OH + 3 O2 → 2 CO2 + 3 H2O How many moles of CO2 can be produced from 7.75 molC2H5OH? 7.75 mol C2H5OH 2 mol CO2= 15.5 mol CO2 1 mol C2H5OH or 2 mol CO2= X = 15.5 mol CO2 1 mol C2H5OH7.75 mol C2H5OH

  4. 4 HCl + O2 → 2 Cl2 + 2 H2O How many moles of Cl2can be produced from 5.60 molHCl? 5.60 mol HCl2 mol Cl2 = 2.80 mol Cl2 4 mol HCl or 2 mol Cl2= X =2.80 mol Cl2 4 mol HCl 5.60 mol HCl

  5. 13. MnO2 + 4 HCl → Cl2 + MnCl2 + 2 H2O How many moles of HCl will react with 1.05 mol MnO2? 1.05 mol MnO24 mol HCl = 4.20 mol HCl 1 mol MnO2 or 1.05 mol MnO2 = 1 mol MnO2 X mol HCl 4 mol HCl X = 4.20 mol HCl

  6. 14. Al4C3 + 12 H2O → 4 Al(OH)3 + 3 CH4 • How many moles of water are needed to react with 100. g Al4C3 ? • How many moles of Al(OH)3 will be produced when 0.600 mol • of CH4 is formed? 100.0 g Al4C31 molAl4C312 mol H2O = 8.333 mol H2O 144.0g Al4C3 1 molAl4C3 0.600 mol CH4 4 mol Al(OH)3 = 0.800 mol Al(OH)3 3 mol CH4

  7. 15. Ca(OH)2 + Na2CO3→ 2 NaOH+ CaCO3 How many grams of sodium hydroxide can be produced from 500 g of calcium hydroxide according to this equation? 1 mole Ca(OH) 2= 74.1 g Ca(OH) 2= 500 g 2 moles NaOH 80.0 g NaOHx 74.1 x = 40000 x = 540 g NaOH

  8. How many grams of zinc phosphate, Zn3(PO4)2, are formed when 10.0 g of Zn are reacted with phosphoric acid? 3 Zn + 2 H3PO4 → Zn3(PO4)2 + 3 H2 10.0 g ? g 3 mol Zn 1 mol Zn3(PO4)2 = 196.2 g Zn 386.1 g Zn3(PO4)2 196.2 g Zn = 10.0 g Zn 386.1 g Zn3(PO4)2 X g Zn3(PO4)2 X = 19.7 gZn3(PO4)2

  9. 17. Fe2O3+ 3C → 2Fe + CO How many kilograms of iron would be formed from 125 kg of Fe2O3? 1 mole Fe2O3 = 0.1598 kg= 125 kg 2 mole Fe 0.1118 kg x 0.1598 x = 13.975 x = 13.975 = 87.5 kg Fe 0.1598

  10. 18.How many grams of steam and iron must react to produce 375 g of magnetic iron oxide, Fe3O4? 3 Fe + 4 H2O → Fe3O4 + 4 H2 3 mol Fe 1 mol Fe3O4 = 167.4 g Fe 231.6 g Fe3O4 = X g Fe 375 g Fe3O4 X = 271 g Fe 4 mol H2O 1 mol Fe3O4 = 72.0 g H2O 231.6 g Fe3O4 = X g H2O 375 g Fe3O4 X = 117 g H2O

  11. 19. a. 2C2H6+ 7O2→ 4CO2 + 6H2O How many moles of O2 are needed for the complete combustion of 15.0 mol of ethane? 7 mole O2= X mole O2 2 mole C2H6 15 mole C2H6 2x = 105 x = 52.5 mole O2

  12. 19 b. 2C2H6+ 7O2→ 4CO2 + 6H2O How many grams of CO2are produced for each 8.00 g of H2O produced? 4 mole CO2 = 176 g CO2 = Xg CO2 6 mole H2O 108 g H2O 8.00gH2O 108x = 1408 x = 1408 = 13.0 g CO2 108

  13. 19 c. 2C2H6+ 7O2 → 4CO2 + 6H2O How many grams of CO2will be produced by the combustion of 75.0 g of C2H6? 4 mole CO2 = 2 = 88.0 g CO2 = x 2 mole C2H6 1 30.0g C2H6 75.0 30x = 6600 x = 6600 = 220. g CO2 30

  14. 20. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 How many moles of Fe2O3 can be made from 1.00 mol of FeS2 ? 4 mol FeS2 2 mol Fe2O3 = 1.00 mol FeS2 X mol Fe2O3 X = 0.500 mol Fe2O3

  15. 20. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (b) How many moles of O2 are required to react with 4.50 mol of FeS2 ? 4 mol FeS2 11 mol O2 = 4.50 mol FeS2 X mol O2 X = 12.4 mol O2

  16. 20. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (c) If the reaction produces 1.55 mol of Fe2O3 , how many moles of SO2 are produced? 2 mol Fe2O3 8 mol SO2 = 1.55 mol Fe2O3 X mol SO2 X = 6.20 mol SO2

  17. 20. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (d) How many grams of SO2 can be formed from 0.512 mol of FeS2 ? 0.512 mol FeS2 8 mol SO2 4 mol FeS2 = 1.02 mol SO2 1.02 mol SO2 64.1 g SO2 1 mol SO2 = 65.4 g SO2

  18. 20. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (e) If the reaction produces 40.6 g of SO2 , how many moles of O2 were reacted? 40.6 g SO2 1 mol SO2 64.1 g SO2 = 0.633 mol SO2 0.633 mol SO2 11 mol O2 8 mol SO2 = 0.870 mol O2

  19. 20. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (f) How many grams of FeS2 are needed to produce 221 g of Fe2O3 ? 221 g Fe2O3 1 mol Fe2O3 159.7 g Fe2O3 = 1.38 mol Fe2O3 1.38 mol Fe2O3 4 mol FeS2 2 mol Fe2O3 = 2.76 mol FeS2 2.76 mol FeS2 120.0 g FeS2 1 mol FeS2 = 331 g FeS2

  20. 20. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (f) How many grams of FeS2 are needed to produce 221 g of Fe2O3 ? 4 mol FeS2 2 mol Fe2O3 = 480.0 g FeS2 319.4 g Fe2O3 = X g FeS2 221 g Fe2O3 X = 332 g FeS2

  21. 22. a) 2 Bi(NO3)3 + 3 H2S → Bi2S3 + 6 HNO3 50.0 g 6.00 g ? g 2 Bi(NO3)3 = 790 g Bi(NO3)3 = 50.0 g Bi(NO3)3 1 Bi2S3 514.3 g Bi2S3 X grams Bi2S3 X = 32.6 grams 3 H2S = 102.3 g H2S = 6.00 g H2S 1 Bi2S3 514.3 g Bi2S3 X g Bi2S3 H2S limits X = 30.2 grams

  22. 22. b.) 3 Fe + 4 H2O → Fe3O4 + 4 H2 40.0 g 16.0 g ? grams 3Fe = 167.4 g Fe = 40.0 g Fe x= 1.91 g H2 4H2 8.0 g x g H2 4H2O = 1H2O = 18.0g H2O = 16.0 g H2O 4H2 1H2 2.0 g H2 x g H2 H2O limiting x= 1.78 g H2

  23. 23) a)C3H8+ 5 O2 → 3 CO2 + 4 H2O 20.0 g 20.0 g ? mol 20.0 g C3H8 1 mol C3H8 3 mol CO2 = 1.36 mol CO2 44.0 g C3H8 1 mol C3H8 20.0 g O2 1 mol O2 3 mol CO2= 0.375 mol CO2 32.0 g O2 5 mol O2

  24. 23 b) C3H8+ 5 O2 → 3 CO2 + 4 H2O 20.0 g 80.0 g ? mol 20.0 g C3H8 1 mol C3H8 3 mol CO2 = 1.36 mol CO2 44.0 g C3H8 1 mol C3H8 80.0 g O2 1 mol O2 3 mol CO2 = 1.50 mol CO2 32.0 g O2 5 mol O2

  25. 24. C3H8 + 5O2 → 3CO2 + 4H2O 5.0 mole 5.0 mole ? Moles 1C3H8 = 5C3H8 X = 15 mole CO2 3 CO2 X 5O2 = 5 X = 3.0 mole CO2 3CO2 X O2 limits

  26. 24. C3H8 + 5O2 → 3CO2 + 4H2O 3.0 mole 20. mole ? moles b) 1 mole C3H8 = 3.0 C3H8 3 mole CO2 X X = 9.00 mole CO2 C3H8 limits 5 mole O2 = 20. O2 3 mole CO2 X X = 12 mole CO2

  27. 24. C3H8 + 5O2 → 3CO2 + 4H2O 3.0 mole 20. mole ? moles 24c) 1 C3H8 = 20 C3H8 3 CO2 X X = 60 mol CO2 5 O2 = 3 O2 3 CO2 X X = 1.8 mol CO2 O2 limits

  28. 27) 2 Al + 3 Br2 → 2 AlBr3 25.0 g 100.0 g ? g If 64.2 g of AlBr3 is recovered, what is the percent yield? 2 mol Al = 54.0 g Al = 25.0 g Al X = 247 g 2 mol AlBr3 533.4 g AlBr3 X g AlBr3 3 mol Br2 = 479.4 g Br2 = 100.0 g Br2 X = 111.3 g 2 mole AlBr3 533.4 g AlBr3 X g AlBr3 % yield = Actual (lab) x 100 = 64.2 g = 57.7 % Calculated (theory) 111.3 g

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