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# 17 stoichiometry

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• ### Transcript of "17 stoichiometry"

1. 1. STOICHIOMETRY
2. 2. STOICHIOMETRY 1 molecule of O2 2H2 + O2  2H2O2 molecules of H2 2 molecules of H2O Now that you know the mole… 1 mole of O2 2H2 + O2  2H2O 2 moles of H2 2 moles of H2O
3. 3. STOICHIOMETRY 1 mole of O2 2H2 + O2  2H2O 2 moles of H2 2 moles of H2O Think of a reaction as a ratio… 2 moles H2 : 1 mole O2 : 2 moles H2O This is called a mole ratioMultiply by 2 = 4 moles H2 : 2 mole O2 : 4 moles H2O
4. 4. STOICHIOMETRY 2 moles H2 : 1 mole O2 : 2 moles H2O How many molecules of water are formed when 3.5 moles of O2 react with H2?= [2 moles H2 : 1 mole O2 : 2 moles H2O] x 3.5= 7 moles H2 : 3.5 moles O2 : 7 moles H2O 7 moles of waterN = n x NA = 7 mol water x 6.02x1023 molecules/mol = 4.21x1024 molecules of water Therefore 4.21x1024 molecules of water are formed.
5. 5. STOICHIOMETRY C2H6 + O2  CO2 + H2OA) How many moles of O2 are required toreact with 13.9mol of C2H6 (ethane)?B)What volume of H2O would be producedby 1.40 mol of O2 and sufficient ethane
6. 6. STOICHIOMETRY C2H6 + O2  CO2 + H2OShould always balance the equation first 2C2H6 + 7O2  4CO2 + 6H2O
7. 7. STOICHIOMETRY 2C2H6 + 7O2  4CO2 + 6H2OA) How many moles of O2 are required to react with 13.9mol of C2H6 (ethane)?iven: mole ratio of C2H6:O2 = 2:7 moles of C2H6 = 13.9mol 2 mol C2H6 = 13.9 mol C2H6 7 mol O2 x 0.2857 = 13.9 x x = 13.9 0.2857 = 48.6 mol of O2 Therefore 48.6 mol of O2 are required
8. 8. STOICHIOMETRY 2C2H6 + 7O2  4CO2 + 6H2OWhat volume of H2O would be produced by 1.40 mol of O2 and sufficient ethaneiven: mole ratio of O2:H2O = 7:6 moles of O2 = 1.40mol 6 mol H2O = x 7 mol O2 1.40 mol O2 0.8571 = x 1.40 x = 0.8571 x 1.40 = 1.20 mol of H2O But the question asked for volume!!!
9. 9. STOICHIOMETRY 2C2H6 + 7O2  4CO2 + 6H2OWhat volume of H2O would be produced by 1.40 mol of O2 and sufficient ethane x = 1.20 mol of H2O Given: n = 1.20 mol MH2O = 18.016 g/mol m=? m = nxM = 1.20 mol x 18.016 g/mol = 21.6 g 1 gram of water = 1mL at room temperature Therefore the volume of water produced is 21.6mL
10. 10. STOICHIOMETRYGeneral rules for solving problems…STEP 1: Write a balanced chemical equationSTEP 2: If you’re given m or N of a substance, convert it to the number of molesSTEP 3: Calculate the number of moles of the required substance based on the number of moles of the given substance (using the appropriate mole ratio)STEP 4: Convert the number of moles of the required substance to mass or number of particles, as directed by the question
11. 11. STOICHIOMETRYPassing chlorine gas through moltensulfur produces liquid disulfur dichloride.How many molecules of chlorine react toproduce 50.0g of disulfur dichloride?
12. 12. STOICHIOMETRYPassing chlorine gas through molten sulfur produces liquiddisulfur dichloride. How many molecules of chlorine react toproduce 50.0g of disulfur dichloride?STEP 1: Write the balanced equation2S + Cl2  S2Cl2STEP 2: Convert the given mass of disulfur dichlorideto the number of molesMS2Cl2 = 135.04g/moln = m/M = 50.0g/135.04g/mol = 0.37026mol
13. 13. STOICHIOMETRYPassing chlorine gas through molten sulfur produces liquiddisulfur dichloride. How many molecules of chlorine react toproduce 50.0g of disulfur dichloride?STEP 3: Calculate the number of moles of the requiredsubstance using your mole ratio2S + Cl2  S2Cl2Given: mole ratio of S2Cl2:Cl2 = 1:1 1 mol S2Cl2 = 0.37026 mol S2Cl2 1 mol Cl2 x x = 0.37026 mol Cl2
14. 14. STOICHIOMETRY Passing chlorine gas through molten sulfur produces liquid disulfur dichloride. How many molecules of chlorine react to produce 50.0g of disulfur dichloride?STEP 4: Convert the number of moles of chlorine gas to thenumber of particlesx = 0.37026 mol Cl2N = n x NA = 0.37026mol x 6.02x1023 molecules/mol = 2.23 x 1023 moleculesTherefore the number of molecules of chlorine is 2.23 x 10 23
15. 15. STOICHIOMETRYTHE LIMITING REACTANT…
16. 16. STOICHIOMETRY THE LIMITING REACTANT… 2C2H6 + 7O2  4CO2 + 6H2ONormally, we assume that all reactants areconsumed in a reaction…Reactants are said to be in stoichiometicamounts when all reactants are consumed inthe ratios predicted
17. 17. STOICHIOMETRY THE LIMITING REACTANT… 2C2H6 + 7O2  4CO2 + 6H2OBut there are often reactants that remain“unreacted”… O2 O2 O2 O2 O2 GAS GAS
18. 18. STOICHIOMETRY THE LIMITING REACTANT…Lithium nitride reacts with water toform ammonia and lithium hydroxide.If 4.87g of lithium nitride reacts with5.80g of water, find the limitingreactant.
19. 19. STOICHIOMETRY THE LIMITING REACTANT… Lithium nitride reacts with water to form ammonia and lithium hydroxide. If 4.87g of lithium nitride reacts with 5.80g of water, find the limiting reactant.STEP 1: Write a balanced chemical equation Li3N + 3H2O  NH3 + 3LiOH
20. 20. STOICHIOMETRY THE LIMITING REACTANT…Lithium nitride reacts with water to form ammonia and lithium hydroxide.If 4.87g of lithium nitride reacts with 5.80g of water, find the limitingreactant. STEP 1: Write a balanced chemical equation Li3N + 3H2O  NH3 + 3LiOH STEP 2: Convert the given masses to the number of moles nLi3N = m/M = 4.87g/34.8g/mol = 0.140mol nH2O = m/M = 5.80g/18.0g/mol = 0.322mol
21. 21. STOICHIOMETRY THE LIMITING REACTANT…Lithium nitride reacts with water to form ammonia and lithium hydroxide.If 4.87g of lithium nitride reacts with 5.80g of water, find the limitingreactant. Li3N + 3H2O  NH3 + 3LiOHnLi3N = 0.140molnH2O = 0.322mol STEP 3: Calculate the number of moles of NH3 produced by both amounts of reactants 1 mol Li3N = 0.140 mol Li3N 1 mol NH3 x x = 0.140 mol NH3
22. 22. STOICHIOMETRY THE LIMITING REACTANT…Lithium nitride reacts with water to form ammonia and lithium hydroxide.If 4.87g of lithium nitride reacts with 5.80g of water, find the limitingreactant. Li3N + 3H2O  NH3 + 3LiOHnLi3N = 0.140molnH2O = 0.322mol STEP 3: Calculate the number of moles of NH3 produced by both amounts of reactants 3 mol H2O = 0.322 mol H2O 1 mol NH3 x x = 0.107 mol NH3 5.80g water makes less ammonia than 4.87g lithium nitride
23. 23. STOICHIOMETRY THE LIMITING REACTANT…Lithium nitride reacts with water to form ammonia and lithium hydroxide.If 4.87g of lithium nitride reacts with 5.80g of water, find the limitingreactant. Li3N + 3H2O  NH3 + 3LiOH Therefore water is the limiting reactant
24. 24. STOICHIOMETRY THE LIMITING REACTANT… P4 + O2  P4O10 A 1.00g piece of phosphorus is burned in aflask filled with 2.60x1023 molecules of oxygen gas. What mass of tetraphosphorus decaoxide is produced?
25. 25. STOICHIOMETRY THE LIMITING REACTANT…A 1.00g piece of phosphorus is burned in a flask filled with 2.60x1023 molecules ofoxygen gas. What mass of tetraphosphorus decaoxide is produced?STEP 1: P4 + 5O2  P4O10STEP 2: nP = 1.00g P4 no = 2.60x1023 molecules 123.9g/mol P4 6.02x1023molecules/mol = 8.07x10-3 mol = 0.432 mol O2 LIMITING REACTANTSTEP 3: Amount of P4O10 produced by P4 = 8.07x10-3 mol Amount of P4O10 produced by O2 = 0.432 ÷ 5 = 8.64 x 10-2 mol
26. 26. STOICHIOMETRY THE LIMITING REACTANT…A 1.00g piece of phosphorus is burned in a flask filled with 2.60x1023 molecules ofoxygen gas. What mass of tetraphosphorus decaoxide is produced?Amount of P4O10 produced by P4 = 8.07x10-3 molm= n x M = 0.00807mol x 284g/mol P4O10 = 2.29g P4O10 Therefore 2.29g of P4O10 is produced.
27. 27. STOICHIOMETRY THE LIMITING REACTANT…empty case battery circuitry iphone 4 1 + 1 + 1  1
28. 28. STOICHIOMETRY THE LIMITING REACTANT…empty case battery circuitry iphone 4What if you were given… Limiting ingredient 3 + 6 + 1  1 ? How many iphones can you make?
29. 29. STOICHIOMETRY THE LIMITING REACTANT…empty case battery circuitry iphone 4What if you were given… 101 + 64 + 97  64 ? Limiting ingredient How many iphones can you make?