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Chapter 11 Stoichiometry

Chapter 11 Stoichiometry. Stoichiometry. Stoichiometry is the study of the , or the measurable, relationships that exist in chemical formulas and chemical reactions. 11-2. Mole to Mole Problems.

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Chapter 11 Stoichiometry

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  1. Chapter 11 Stoichiometry

  2. Stoichiometry Stoichiometry is the study of the, or the measurable, relationships that exist in chemical formulas and chemical reactions. 11-2

  3. Mole to Mole Problems Molar ratios (ratios of the in a balanced equation) can be used to determine the number of moles of in the reaction if you know the number of moles of other substance in the reaction. 11-3

  4. Mole to Mole Problems Ex. How many moles of Al(NO3)3 are produced when mol of AgNO3 react in this equation? 3AgNO3 + Al → Al(NO3)3 + 3Ag 0.75mol AgNO3 x 1mol Al(NO3)3=mol Al(NO3)3 3mol AgNO3 11-4

  5. Mole to Mole Problems 1) Now you try one... will react with to produce Fe2O3. How many moles of Fe2O3 will be produced if 0.80 mol of Fe reacts? First step, figure out the equation~ Fe + O2→ Fe2O3 11-5

  6. Mole to Mole Problems Now balance the equation ~ Fe + O2→ Fe2O3 Now set up your proportion: 0.80 mol Fe x 2 mol Fe2O3= mol Fe2O3 4 mol Fe 11-6

  7. Mass to Mass Problems That was a piece of cake… Let’s try some HARDER problems…

  8. Mass to Mass Problems Given the of one substance, find the of another involved the same reaction. 11-8

  9. Mass to Mass Problems

  10. Mass to Mass Problems Ex. How much NaOH is made when 0.25g of Na reacts with H2O? First, write a balanced equation: Na + H2O → NaOH + H2 Na + H2O → NaOH + H2 11-10

  11. Mass to Mass Problems Ex. How much NaOH is made when 0.25g of Na reacts with H2O? 2Na + 2H2O → 2NaOH + H2 Now the given mass to : 0.25g Na x 1mol Na = mols Na 22.99 g Na Next, determine the # of mols of NaOH by using the : 0.011 mols Na x 2 mol NaOH = molNaOH mol Na Last step, convert to moles using molar mass: mol NaOH x 39.99 g/mol NaOH = g NaOH

  12. Mass to Mass Problems Let’s try another, carrying out the steps all in : Ex. How much LiOH is produced when 0.38g of Li3N reacts with H2O? Li3N + H2O → NH3 + LiOH Li3N + H2O → NH3 + LiOH 0.38gLi3N x 1mol Li3N x 3mol LiOH x 23.94g LiOH 34.83g Li3N 1molLi3N 1mol LiOH = g LiOH 11-12

  13. Mass to Mass Problems 2) Now you try one: What mass of sodium chloride is produced when chlorine reacts with 0.30g of sodium iodide? NaI + Cl2 → I2 + NaCl NaI + Cl2 → I2 + NaCl 0.30gNaI x 1mol NaI x 2mol NaCl x 58.44g NaCl 149.89g NaI 2molNaI 1mol NaCl = g NaCl 11-13

  14. Mass ~ Volume Problems Given the of one substance in a reaction, find the of another substance. Ex. How many of CO2 are produced at when 400.00g of CaCO3 react with HCl? 11-14

  15. Mass ~ Volume Problems Ex. How many liters of CO2 are produced at STP when 400.00g of CaCO3 react with HCl? CaCO3 + HCl → CaCl2 + CO2 + H2O CaCO3 + HCl → CaCl2 + CO2 + H2O 400.00g CaCO3 x 1mol CaCO3 x 1mol CO2 x 22.4L 100.04g CaCO3 1mol CaCO3 1mol CO2 = L CO2 11-15

  16. Mass ~ Volume Problems 3) Now you try one: How many liters of oxygen are necessary for the combustion of 134g of Mg at STP? 2Mg + O2 → 2MgO 134.00g Mg x 1mol Mg x 1mol O2 x 22.4L g Mg 2mol Mg 1 mol CO2 = L O2 11-16

  17. Mass ~ Volume Problems You can also work these problems in reverse. Given the produced, find the. 11-17

  18. Volume ~ Mass Problems Ex. Find the mass of required to produce 1.82L of CO2 at in the rxn: C6H12O6 → 2C2H6O + 2 CO2 1.82L CO2 x 1mol CO2 x 1molC2H12O6 x 180gC6H12O6 22.4L 2mol CO2 1 mol = g of C6H12O6 11-18

  19. Volume ~ Volume Problems Given a volume, find a volume: Ex. When 0.75 L of reacts with, what volume of HBr is produced? H2 + Br2 → 2HBr 0.75L H2 x 2L HBr =1.5L HBr 1L H2 11-19

  20. Volume ~ Volume Problems 4) What volumes of sulfur dioxide and dihydrogen sulfide gases are necessary to produce 11.4 L of water vapor? SO2 + H2S → S + H2O 11.4 L H2O x 1L SO2 = L SO2 2L H2O 11.4L H2O x 2L H2S = L H2S 2L H2O

  21. It’s Packet Time!! Try pages 8, 10, 13 & 14 in your packets covering mole to mole problems, mass to mass problems, mass to volume problems and volume to volume problems. 11-21

  22. Limiting Reactants The limiting reactant is the reactant that limits the of formed in a chemical reaction. (The of product is always determined by the quantity of the limiting reactant.) 11-22

  23. Limiting Reactants When doing limiting reactant problems, you do a calculation for each reactant with the product. The reactant that produces the amount of is the limiting reactant. 11-23

  24. Limiting Reactants Ex. Identify the limiting reactant when 1.22g of O2 reacts with 1.05g H2 to form water. H2 + O2 → H2O 1.22gO2 x 1mol O2 x 2mol H2O x 18.02g H2O 32.00g O2 1 mol O2 1mol H2O = g H2O 1.05gH2 x 1mol H2 x 2mol H2O x 18.02gH2O 2.02g H2 2mol H2 1mol H2O = g H2O is the limiting reactant.

  25. Limiting Reactants Ex. If 4.10g of Cr is heated with 9.30g Cl2, what mass of CrCl3 is produced? Cr + Cl2 → CrCl3 4.10gCr x 1mol Cr x 2mol CrCl3 x 158.35gCrCl3 52.0g Cr 2 mol Cr1mol H2O = g CrCl3 9.30gCl2 x 1mol Cl2 x 2mol CrCl3 x158.35gCrCl3 70.9g H2 3mol Cl2 1mol CrCl3 = g CrCl3 g CrCl3 is produced.

  26. Limiting Reactants 5) What mass of SO3 is produced from the reaction of 12.4g SO2 and 3.45g O2? SO2 + O2 → SO3 12.4gSO2 x 1mol SO2 x 2mol SO2 x 80.07g 64.07g SO2 2 mol SO3 1mol SO3 = g SO3 3.45gO2 x 1mol O2 x 2mol SO3 x 80.07gSO3 32.0g O2 1mol O2 1mol SO3 = g SO3 g SO3 is produced.

  27. Percent Yield yield is the percent of the expected yield that was actually obtained. yield is the amount of product that should be produced based on calculations. Yield is the percent of the expected yield that was actually obtained. % Yield = Actual Yield x 100% Expected Yield 11-27

  28. Percent Yield % Yield = Actual Yield x 100% Yield Ex. Determine the percent yield for the reaction between 6.92g of and 4.28g of if 7.36 g of KO2 is produced. First find the limiting reactant K + O2→KO2 11-28

  29. Percent Yield Ex. Determine the percent yield for the reaction between 6.92g of K and 4.28g of O2 if 7.36 g of KO2 is produced. First find the limiting reactant: K + O2→KO2 6.92g K x 1mol K x 1mol KO2 x 71.1g = g 39.1g K 1 mol K 1mol KO2 KO2 4.28g O2 x 1mol O2x 1mol KO2x 71.1g = g 32.0g O2 1mol O2 1mol KO2 KO2 is the limiting reactant, the expected yield of KO2 is g. % Yield = 7.36g KO2 x 100 = % 9.51g KO2

  30. Percent Yield 6) Determine the percent yield for the reaction between 45.9g of NaBr and excess chlorine gas to produce 12.8g NaCl and and unknown quantity of bromine gas. NaBr + Cl2→ NaCl + Br2 Expected Yield = 45.9g NaBr x 1mol NaBr x 2mol NaCl x 58.44g = 102.89g NaBr 2 mol NaBr 1mol NaCl The expected yield of NaCl is g . % Yield = 12.8g NaCl x 100 = % g NaCl 11-30

  31. Packet Time #2 Now try pages 19 & 20 in your Ch. 11 packets with practice problems on limiting reactants and % yield. 11-31

  32. Chapter 11 The End!

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