Stoichiometry Mass Changes in Chemical Reactions - PowerPoint Presentation

1. StoichiometryMass Changes in Chemical ReactionsLimiting reactantsPercentage yield

2. Stoichiometry ProblemsHow many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? = 6 mol KClO3Problem: X molKClO3  9mol O22KClO3  2KCl + 3O2 Balanced : 2 molKClO3  3mol O2EquationExample 1

3. Stoichiometry ProblemsHow many grams of silver will be formed when 12 g of copper reacts with aluminum nitrate to produce silver and copper II nitrate and silver? = 41 g AgProblem: 12gCu  Xg AgCu + 2 AgNO3  2 Ag + Cu(NO3)2 Balanced: 63.5 gCu  2(107.9) g AgEquation 215.8 gExample2

4. Stoichiometry ProblemsIf 12.0 grams of potassium chlorate decompose, how many moles of potassium chloride will be produced?= 0.0988 mol KClProblem: 12gKClO3  X moles KCl2 KClO3  2KCl + 3 O2Balanced: 2(122.6) g KClO3  2 moles KClEquation 245.2 gExample3

5. Stoichiometry ProblemsIn an experiment, red mercury (II) oxide powder is placed in an open flask and heated until it is converted to liquid mercury and oxygen gas. The liquid mercury has a mass of 92.6 g. What is the mass of oxygen formed in the reaction? = 7.39 g O2Problem: 92.6 g Hg + X g O22HgO  2Hg + O2Balanced: 2 ( 200.6) g Hg + 32 g O2Equation 401.2 g HgLEARNINGCHECK

6. Limiting Reactant

7. Bike AnalogyConsider the following Analogy:2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bikeHow many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain?

8. Bike AnalogyLimiting ReactantExcess ReactantConsider the following Analogy:2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bikeHow many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain?

9. Cheeseburger AnalogyConsider the following Analogy:2 Cheese + 1 burger patty + 1 bun = cheese burgerGiven three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made?

10. Cheeseburger AnalogyConsider the following Analogy:2 Cheese + 1 burger patty + 1 bun = 1 cheese burgerGiven three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made?LRER

11. Limiting Reactant vs. Excess ReactantsLimiting reactant is the reactant that runs out firstWhen the limiting reactant is exhausted, then the reaction stops In our examples, the limiting reactants will be the wheels in the bicycle analogy and the burger patty in our hamburger analogy

12. 1. Write a balanced equation.2. For each reactant, calculate the amount of product formed.3. The reactant that resulted in the smallest amount of product is the limiting reactant(LR).4. To find the amount of leftover reactant—excess—calculate the amount of the no LR used by the LR.5. Subtract the calculated amount in step 4 from the original no LR amount given in the problem.Limiting Reactants Calculations

13. Example 1Determine how many moles of water can be formed if I start with 2.75 moles of hydrogen and 1.75 moles of oxygen.= 2.75 mol H2OProblem: 2.75 mol H2  XmolH2O2H2 + O2  2H2O Balanced: 2 mol H2  2molH2OEquation= 3.50 mol H2OProblem: 1.75 mol O2  XmolH2O2H2 + O2  2H2O Balanced: 1 mol O2  2molH2OEquationLimiting reactant =H2

14. If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant?= 1.0 mol H2OProblem: 2.0 mol HF  XmolH2OSiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)Balanced: 4 mol HF  2molH2OEquation= 9.0 mol H2OProblem: 4.5 mol SiO2  XmolH2OSiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)Balanced: 1 mol SiO2  2molH2OEquationLimiting reactant =HFExample2

15. If 36.0 g of H2O is mixed with 167 g of Fe , which is the limiting reactant?= 106 g Fe2O3Problem: 36.0 g H2O  XgFe2O32Fe(s) + 3H2O(g)  Fe2O3(g) + 3H2(g)Balanced: 54 g H2O  159.6gFe2O3EquationLimiting reactant =H2O= 238 g Fe2O3Problem: 167 g Fe  XgFe2O32Fe(s) + 3H2O(g)  Fe2O3(g) + 3H2(g)Balanced: 111.6 g Fe  159.6gFe2O3EquationLEARNINGCHECK

16. 1. Write a balanced equation.2. For each reactant, calculate the amount of product formed.3. The reactant that resulted in the smallest amount of product is the limiting reactant(LR).4. To find the amount of leftover reactant—excess—calculate the amount of the no LR used by the LR.5. Subtract the calculated amount in step 4 from the original no LR amount given in the problem.Limiting Reactants Calculations

17. Limiting ReactantsLimiting reactant: H2OExcess reactant: FeProducts Formed: 107 g Fe3O3 & 4.00 g H2 left over iron 3Fe(s) + 4H2O(g) Fe3O3(g) + 4H2(g)LRXS3Fe(s) + 4H2O(g) Fe3O3(g) + 4H2(g)Problem: XgFe 36.0 g H2OBalanced: 111.6 g Fe 54 g H2OEquation= 74.4 g Fe used167gFe - 74.4 g Fe= 92.6 g FeOriginal – Used = Excess

18. So far, the masses we have calculated from chemical equations were based on the assumption that each reaction occurred 100%.The THEORETICAL YIELD the maximum amount of product that can be produced in a reaction (calculated from the balanced equation)The ACTUAL YIELD is the amount of product that is “actually” produced in an experiment (usually less than the theoretical yield)Percent Yield

19. Percent Yield Theoretical Yield the maximum amount of product that can be produced in a reactionPercent Yield The actual amount of a given product as the percentage of the theoretical yield.

20. Look back at the problem from LEARNING CHECK. We found that 106 g Fe2O3 could be formed from the reactants.In an experiment, you formed 90.4 g of Fe2O3. What is your percent yield?% Yield = 90.4 g x 100 = 85.3% 106 g

21. A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this?= 19.1 g CH3COOC2H5Problem: 10.0g C2H5OH  Xg CH3COOC2H5CH3COOH + C2H5OH  CH3COOC2H5 + H2OBalanced: 46.0 g C2H5OH  88.0 g CH3COOC2H5Equation% Yield = 14.8 g x 100 = 77.5% 19.1gExample1

22. When 36.8 g of C6H6 reacts with Cl2, what is thetheoretical yield of C6H5Cl produced? If the actual is43.7 g, determine the percentage yield of C6H5Cl.= 53.1 g C6H5ClProblem: 36.8g C5H5  Xg C5H5Cl2C6H6 + Cl2  2C6H5Cl + H2Balanced: 156.0 g C5H5  225.0 g C5H5ClEquation% Yield = 43.7 g x 100 = 88.3% 53.1gLEARNINGCHECK