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Stoichiometry Mass Changes in Chemical Reactions

Stoichiometry Mass Changes in Chemical Reactions. Limiting reactants Percentage yield. Stoichiometry Problems. Example 1. How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? . Problem: X mol KClO 3  9mol O 2.

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Stoichiometry Mass Changes in Chemical Reactions

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  1. StoichiometryMass Changes in Chemical Reactions Limiting reactants Percentage yield

  2. Stoichiometry Problems Example 1 How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? Problem: X molKClO3 9mol O2 2KClO3 2KCl + 3O2 Balanced : 2 molKClO3 3mol O2 Equation = 6 mol KClO3

  3. Stoichiometry Problems Example 2 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? Problem: 12gCu  Xg Ag Cu + 2 AgNO3 2 Ag + Cu(NO3)2 Balanced: 63.5 gCu 2(107.9) g Ag Equation 215.8 g = 41 g Ag

  4. Stoichiometry Problems Example 3 • If 12.0 grams of potassium chlorate decompose, how many moles of potassium chloride will be produced? Problem: 12gKClO3 X moles KCl 2 KClO3  2KCl + 3 O2 Balanced: 2(122.6) g KClO3 2 moles KCl Equation 245.2 g = 0.0988 molKCl

  5. Stoichiometry Problems LEARNING CHECK • 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? Problem: 92.6 g Hg + X g O2 2HgO  2Hg + O2 Balanced: 2 ( 200.6) g Hg + 32 g O2 Equation 401.2 g Hg = 7.39 g O2

  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 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? Limiting Reactant Excess Reactant

  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 LR 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? 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. Limiting Reactants Calculations 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.

  13. Example 1 Problem: 2.75 mol H2 XmolH2O Determine how many moles of water can be formed if I start with 2.75 moles of hydrogen and 1.75 moles of oxygen. 2H2 + O2  2H2O Limiting reactant =H2 Balanced: 2 mol H2 2molH2O Equation = 2.75 mol H2O Problem: 1.75 mol O2 XmolH2O 2H2 + O2  2H2O Balanced: 1 molO2 2molH2O Equation = 3.50 mol H2O

  14. Example2 Problem: 2.0 mol HF  XmolH2O • If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant? SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l) Balanced: 4 molHF  2molH2O Equation = 1.0 mol H2O Limiting reactant =HF Problem: 4.5 molSiO2  XmolH2O SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l) Balanced: 1 molSiO2 2molH2O Equation = 9.0 mol H2O

  15. LEARNINGCHECK If 36.0 g of H2O is mixed with 167 g of Fe , which is the limiting reactant? Problem: 36.0 g H2O XgFe2O3 2Fe(s) + 3H2O(g)  Fe2O3(g) + 3H2(g) Balanced: 54 g H2O 159.6gFe2O3 Equation = 106 g Fe2O3 Limiting reactant =H2O Problem: 167 g Fe XgFe2O3 2Fe(s) + 3H2O(g)  Fe2O3(g) + 3H2(g) Balanced: 111.6 g Fe 159.6gFe2O3 Equation = 238 g Fe2O3

  16. Limiting Reactants Calculations 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.

  17. left over iron Limiting Reactants XS LR 3Fe(s) + 4H2O(g) Fe3O3(g) + 4H2(g) Limiting reactant: H2O Excess reactant: Fe Products Formed: 107 g Fe3O3 & 4.00 g H2 Problem: XgFe 36.0 g H2O 3Fe(s) + 4H2O(g) Fe3O3(g) + 4H2(g) Balanced: 111.6 g Fe 54 g H2O Equation = 74.4 g Fe used 167gFe - 74.4 g Fe= 92.6 g Fe Original – Used = Excess

  18. Percent Yield • 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)

  19. Percent Yield • 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 Fe2O3could 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. Example1 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? Problem: 10.0gC2H5OH XgCH3COOC2H5 CH3COOH + C2H5OH  CH3COOC2H5+ H2O Balanced: 46.0 gC2H5OH 88.0 g CH3COOC2H5 Equation = 19.1 g CH3COOC2H5 % Yield = 14.8 g x 100 = 77.5% 19.1g

  22. LEARNINGCHECK When 36.8 g of C6H6 reacts with Cl2, what is the theoretical yield of C6H5Cl produced? If the actual is 43.7 g, determine the percentage yield of C6H5Cl. Problem: 36.8gC5H5 XgC5H5Cl 2C6H6+ Cl2 2C6H5Cl + H2 Balanced: 156.0 gC5H5 225.0 g C5H5Cl Equation = 53.1 g C6H5Cl % Yield = 43.7 g x 100 = 88.3% 53.1g

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