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Chapter 12 Stoichiometry

Chapter 12 Stoichiometry. By David B. Oestreicher. What is Stoichiometry?. Stoichiometry is the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction; based on the law of conservation of mass. Desired Learning Objectives.

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Chapter 12 Stoichiometry

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  1. Chapter 12Stoichiometry

    By David B. Oestreicher
  2. What is Stoichiometry? Stoichiometry is the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction; based on the law of conservation of mass.
  3. Desired Learning Objectives You can write mole ratios from balanced chemical equations You can calculate the number of moles and the mass of a reactant or product when given the number of moles or the mass of another reactant or product You can identify the limiting reactant in a chemical reaction You can determine the percent yield of a chemical reaction
  4. Cooking Examples How many eggs will we need to make 3 Western Omelets? How many peppers will we need to make 4 Western Omelets? If we made 40 Western Omelets, how many pieces of cheese did we use? 2 eggs + 1 slice of cheese + 1 pepper1 W. Om
  5. Stoichiometry and balanced equations 4Fe (s) + 3O2------- 2Fe2O3(s) 4 Moles of Fe = 223.4 g of Fe 3 Moles of O2 = 96.00 g of O2 2 Moles Fe2O3 = 319.4 g 223.4 g + 96.0 g = 319.4 g Law of the Conservation of Mass or One side of the equation equals the other
  6. 2Al + 3Br2 2AlBr3How do all the molecules relate to one another? 2 moles Al2 moles Al 3 moles Br22 moles AlBr3 3 moles Br23 moles Br2 2 moles Al 2 moles AlBr3 2 moles AlBr32 moles AlBr3 2 moles Al 3 moles Br2
  7. How many moles of carbon dioxide are produced when 10.0 moles of propane are burned in excess oxygen in a gas grill? First, what do we know? What do we need to find?a. We have 10.0 moles of propaneb. We want to find out how much carbon dioxide will be produced. Write a balanced equation for the problem: C3H8 (g) + 5O2-- 3CO2 + 4H2O(g)
  8. Ratios of reactants to each other and the products C3H8 (g) + 5O2-- 3CO2 + 4H2O(g) The ratio of CO2 to C3H8 is 3:1 or 3 moles CO2 = 3 1 mole C3H8 1 So 10.0 moles C3H8 x 3 moles CO2 = 30.0 moles of CO2 1 mole C3H8
  9. Determine the mass of sodium chloride or table salt (NaCl) produced when 1.25 moles of chlorine gas (Cl2) reacts vigorously with sodium. What do we have? What do we want to find out?a. We have 1.25 moles of Cl2 b. We want to find out the mass of NaCl produced 2. Write a balanced equation for the reaction 2Na(s) + Cl2(g)---- 2NaCl(s) What is the ratio of NaCl to Cl2? 2 moles NaCl or 2:1 1 mole Cl2
  10. Solve using ratios 1.25 moles Cl2 x 2 moles NaCl = 2.5 moles NaCl 1 mole Cl2 Is that the answer? NO! We were asked for the mass of NaCl so we need to convert from moles to grams 2.5 moles NaCl x 58.44 g NaCl = 146 g NaCl 1 mole NaCl
  11. Limiting Reactants The reaction between solid white phosphorus and oxygen produces solid tetraphosphorous decoxide (P4O10). This compound is often called diphosphorous pentoxide because its empirical formula is P2O5. a. Determine the mass of tetraphosphorous decoxide formed if 25.0 g of phosphorous (P4) and 50.0 g of oxygen are combined. b. How much of the excess reactant remains after the reaction stops?
  12. How do we solve this? What do we know? mass of phosphorous = 25.0 g P4 mass of oxygen = 50.0 g O2 Write a balanced equation for the reaction P4(s) + 5O2(g) --- P4O10(s)
  13. Determine the number of moles of reactants P4(s) + 5O2(g) --- P4O10(s) 25.0 g P4x 1 mol P4= 0.202 mol P4 123.9g P4 50.0 g O2x 1 mol O2= 1.56 mol O2 32.00 g O2
  14. P4(s) + 5O2(g) --- P4O10(s) Calculate the actual ratio of available moles of O2 and available moles of P4. 1.56 moles O2= 7.72 moles O2 0.202 moles P4 1 mol P4
  15. Determine the mole ratio of the two reactants from the balanced equation P4(s) + 5O2(g) --- P4O10(s) 5 mol O2 1 mol P4 Because 7.72 mole of O2are available, but only 5 moles are needed to react with 1 mole of P4, the O2 is in excess and P4is the limiting reactant.
  16. Use the limiting reactant’s moles to determine how much product will be produced. 0.202 mol P4 x 1 mol P4O10 = 0.202 mol P4O10 1 mol P4 To find the mass 0.202mol P4O10 x 283.9 g P4O10 = 57.3 g P4O10 1 mol P4O10
  17. b. How much of the excess reactant remains after the reaction goes to completion? 0.202 mol P4 x 5 mol O2 = 1.01 mol O2 (needed) 1 mol P4 1.01 mol O2x 32.00 g O2= 32.3 g O2 needed 1 mol O2 50.0 g O2 given – 32.3 g O2 needed = 17.7 g O2 in excess
  18. Percent Yield Percent yield = actual yield (from an experiment)______________ x 100 theoretical yield (from stoichiometric calculations)
  19. Calculating Percent Yield When potassium chromate (K2CrO4) is added to a solution containing 0.500 g of silver nitrate (AgNO3), solid silver chromate (Ag2CrO4) is formed. Determine the theoretical yield of the silver chromate precipitate. If 0.455 g of silver chromate is obtained, calculate the percent yield.
  20. To Solve What do we know? a. mass of silver nitrate = 0.500 g AgNO3 b.percent yield = ? % yield of Ag2CrO4 Write a balanced equation for the reaction 2AgNO3 + K2CrO4 --- Ag2CrO4 + 2KNO3
  21. 2AgNO3 + K2CrO4 --- Ag2CrO4 + 2KNO3 Convert grams to moles of AgNO3 0.500 g AgNO3x1 mol AgNO3= 2.94 x 10-3 mol AgNO3 169.9 g AgNO3 Use the appropriate mole ratio to convert mol AgNO3 to mol Ag2CrO4 2.94 x 10-3molAgNO3 x 1 mol Ag2CrO4 = 1.47 x 10-3 molAg2CrO4 2 mol AgNO3
  22. Calculate the mass of Ag2CrO4 (theoretical yield) by multiplying mol Ag2CrO4 by the molar mass From last slide 2.94 x 10-3molAgNO3 x 1 mol Ag2CrO4 = 1.47 x 10-3 molAg2CrO4 2 mol AgNO3 1.47 x 10-3 molAg2CrO4 x 331.7 gAg2CrO4 = 0.488 gAg2CrO4 1 mol Ag2CrO4 Divide the actual yield by the theoretical yield and multiply times 100 0.455 g Ag2CrO4 x 100 = 93.2 % Ag2CrO4 0.488 g Ag2CrO4
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