1. Stoichiometry Mass Changes in Chemical Reactions Limiting reactants Percentage yield

2. Stoichiometry Problems How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? = 6 mol KClO 3 Problem: X molKClO 3  9mol O 2 2KClO 3  2KCl + 3O 2 Balanced : 2 molKClO 3  3mol O 2 Equation Example 1

3. Stoichiometry Problems How 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 Ag Problem: 12gCu  Xg Ag Cu + 2 AgNO 3  2 Ag + Cu(NO 3 ) 2 Balanced: 63.5 gCu  2(107.9) g Ag Equation 215.8 g Example 2

4. Stoichiometry Problems If 12.0 grams of potassium chlorate decompose, how many moles of potassium chloride will be produced? = 0.0988 mol KCl Problem: 12gKClO 3  X moles KCl 2 KClO 3  2KCl + 3 O 2 Balanced: 2(122.6) g KClO 3  2 moles KCl Equation 245.2 g Example 3

5. Stoichiometry Problems In 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 O 2 Problem: 92.6 g Hg + X g O 2 2HgO  2Hg + O 2 Balanced: 2 ( 200.6) g Hg + 32 g O 2 Equation 401.2 g Hg LEARNING CHECK

6. Limiting Reactant

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

8. Bike Analogy Limiting Reactant Excess Reactant Consider the following Analogy: 2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain?

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

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

11. Limiting Reactant vs. Excess Reactants –Limiting reactant is the reactant that runs out first –When 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 1 Determine 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 H 2 O Problem: 2.75 mol H 2  XmolH 2 O 2H 2 + O 2  2H 2 O Balanced: 2 mol H 2  2molH 2 O Equation = 3.50 mol H 2 O Problem: 1.75 mol O 2  XmolH 2 O 2H 2 + O 2  2H 2 O Balanced: 1 mol O 2  2molH 2 O Equation Limiting reactant =H 2

14. If 2.0 mol of HF are exposed to 4.5 mol of SiO 2, which is the limiting reactant? If 2.0 mol of HF are exposed to 4.5 mol of SiO 2, which is the limiting reactant? = 1.0 mol H 2 O Problem: 2.0 mol HF  XmolH 2 O SiO 2 (s) + 4HF (g) SiF 4 (g) + 2H 2 O (l) SiO 2 (s) + 4HF (g)  SiF 4 (g) + 2H 2 O (l) Balanced: 4 mol HF  2molH 2 O Equation = 9.0 mol H 2 O SiO 2 Problem: 4.5 mol SiO 2  XmolH 2 O SiO 2 (s) + 4HF (g) SiF 4 (g) + 2H 2 O (l) SiO 2 (s) + 4HF (g)  SiF 4 (g) + 2H 2 O (l) SiO 2 Balanced: 1 mol SiO 2  2molH 2 O Equation Limiting reactant =HF Example 2

15. If 36.0 g of H 2 O is mixed with 167 g of Fe, which is the limiting reactant? = 106 g Fe 2 O 3 H 2 O Problem: 36.0 g H 2 O  XgFe 2 O 3 2Fe (s) + 3H 2 O (g) Fe 2 O 3 (g) + 3H 2 (g) 2Fe (s) + 3H 2 O (g)  Fe 2 O 3 (g) + 3H 2 (g) H 2 O Balanced: 54 g H 2 O  159.6gFe 2 O 3 Equation Limiting reactant = H 2 O = 238 g Fe 2 O 3 Fe Problem: 167 g Fe  XgFe 2 O 3 2Fe (s) + 3H 2 O (g) Fe 2 O 3 (g) + 3H 2 (g) 2Fe (s) + 3H 2 O (g)  Fe 2 O 3 (g) + 3H 2 (g) Fe Balanced: 111.6 g Fe  159.6gFe 2 O 3 Equation LEARNING CHECK

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 Reactants Limiting reactant: H 2 O Excess reactant: Fe Products Formed: 107 g Fe 3 O 3 & 4.00 g H 2 left over iron 3Fe(s) + 4H 2 O(g) Fe 3 O 3 (g) + 4H 2 (g) LR XS H 2 O Problem: XgFe 36.0 g H 2 O H 2 O Balanced: 111.6 g Fe 54 g H 2 O Equation = 74.4 g Fe used 167gFe - 74.4 g Fe= 92.6 g Fe Original – 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. Theoretical Yield the maximum amount of product that can be produced in a reaction Percent 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 Fe 2 O 3 could be formed from the reactants. In an experiment, you formed 90.4 g of Fe 2 O 3. What is your percent yield? % Yield = 90.4 g x 100 = 85.3% 106 g

21. A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What percent yield of ethyl acetate is this? = 19.1 g CH 3 COOC 2 H 5 Problem: 10.0g C 2 H 5 OH  Xg CH 3 COOC 2 H 5 CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5 + H 2 O Balanced: 46.0 g C 2 H 5 OH  88.0 g CH 3 COOC 2 H 5 Equation % Yield = 14.8 g x 100 = 77.5% 19.1g Example 1

22. When 36.8 g of C 6 H 6 reacts with Cl 2, what is the theoretical yield of C 6 H 5 Cl produced? If the actual is 43.7 g, determine the percentage yield of C 6 H 5 Cl. = 53.1 g C 6 H 5 Cl Problem: 36.8g C 5 H 5  Xg C 5 H 5 Cl 2C 6 H 6 + Cl 2  2C 6 H 5 Cl + H 2 Balanced: 156.0 g C 5 H 5  225.0 g C 5 H 5 Cl Equation % Yield = 43.7 g x 100 = 88.3% 53.1g LEARNING CHECK