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Chapter 8 Quantities in Chemical Reactions

Chapter 8 Quantities in Chemical Reactions. Tro, 2 nd ed. STOICHIOMETRY: SOME REVIEW. The molar mass of an element is its atomic mass in grams. It contains 6.0221 x 10 23 atoms (Avogadro’s number) of the element.

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Chapter 8 Quantities in Chemical Reactions

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  1. Chapter 8Quantities inChemical Reactions Tro, 2nd ed.

  2. STOICHIOMETRY: SOME REVIEW The molar mass of an element is its atomic mass in grams. It contains 6.0221 x 1023 atoms (Avogadro’s number) of the element. The molar mass of an element or compound is the sum of the atomic masses of all its atoms.

  3. Avogadro’s Number of Particles 6.0221 x 1023 Particles 1 MOLE Molar Mass

  4. How many moles of NaCl are present in 292.215 grams of NaCl? The molar mass of NaCl =58.443 g.

  5. STOICHIOMETRY Consider the combustion of acetylene: 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) 2 molecules 5 molecules 4 molecules 2 molecules 200 “ 500 “ 400 “ 200 “ 12x1023 31x1023 24x1023 12x1023 2 moles 5 moles 4 moles 2 moles Consider each of the formulas as molecules or moles Coefficients represent number of moles “Recipe” just like a cooking recipe: 2 cups of acetylene requires 5 cups of oxygen, etc. When we use the coefficients to relate the chemicals in an equation, we are doing STOICHIOMETRY

  6. STOICHIOMETRY 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) EXAMPLE: If 6 moles of C2H2 are burned, then 15 moles of O2 are needed, 12 moles of CO2 are made and 6 moles of H2O are made How did we know? We used ratios. Practice with 0.102 moles of C2H2: 0.102 mol * 5 mol O2/2 mol C2H2 = 0.255 mol O2 0.102 mol * 4 mol CO2/2 mol C2H2 = 0.204 mol CO2 0.102 mol * 2 mol H2O/2 mol C2H2 = 0.102 mol H2O

  7. Introduction to Stoichiometry:The Mole-Ratio Method Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products. Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction. The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

  8. 1 mol 3 mol 2 mol The Mole-Ratio Method N2 + 3H2 2NH3

  9. 1 mol 3 mol 2 mol The Mole-Ratio Method N2 + 3H2 2NH3

  10. The Mole-Ratio Method The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem. The mole ratio is used in the solution of every type of stoichiometry problem.

  11. The Mole Ratio Method for Stoichiometry 1. Have a balanced chemical equation. 2. Convert the quantity of starting substance to moles (if it is not already moles) 3. Convert the moles of starting substance to moles of desired substance. 4. Convert the moles of desired substance to the units specified in the problem.

  12. In the following reaction how many moles of PbCl2 are formed if 5.000 moles of NaCl react? 2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq)

  13. Mole Ratio 1 mol 5 mol 3 mol 1 mol 5 mol Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4). Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4 Step 2 Moles starting substance: 10.0 mol H2SO4 Step 3 The conversion needed is moles H2SO4 moles H3PO4

  14. Mole-Mass Calculations The object of this type of problem is to calculate the mass of one substance that reacts with or is produced from a given number of moles of another substance in a chemical reaction. If the mass of the starting substance is given, we need to convert it to moles. (Step 2) We use the mole ratio to convert moles of starting substance to moles of desired substance. (Step 3) We can then change moles of desired substance to mass of desired substance if called for by the problem. (Step 4)

  15. Mole Ratio Method 1 Step by Step Calculations Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4. Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4 Step 1 Balance reaction given, with 784 grams of H3PO4. Step 2 Convert grams of H3PO4 to moles of H3PO4. Step 3 Convert moles of H3PO4 to moles of H2SO4 by the mole-ratio method.

  16. The conversion needed is grams H3PO4 moles H3PO4  moles H2SO4 Mole Ratio Method 2 Continuous Calculation Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4 Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4

  17. Mass-Mass Calculations Solving mass-mass stoichiometry problems requires all four steps of the mole-ratio method. You list the four steps here:

  18. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2. N2 + 3 H2 2 NH3 Method 1 Step by Step Calculations Step 2 Convert 112 g of H2 to moles. grams  moles Step 3 Calculate the moles of NH3 by the mole ratio method.

  19. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2. N2 + 3 H2 2 NH3 Step 4 Convert moles NH3 to grams NH3. moles  grams

  20. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2. N2 + 3 H2 2 NH3 Method 2 Continuous Calculation grams H2 moles H2  moles NH3  grams NH3

  21. GROUP PRACTICE: First write the balanced chemical reaction: ammonium phosphate reacts with calcium hydroxide to produce calcium phosphate, ammonia and water. Then answer: a. What weight of calcium hydroxide is need to produce 155 g of calcium phosphate? b. How many molecules of ammonia gas are released? c. What volume of water will be produced, assuming normal conditions? 2 (NH4)3PO4(aq) + 3 Ca(OH)2(aq) Ca3(PO4)2(aq) + 6 NH3(g) + 6 H2O(l)

  22. a. 155 g Ca3(PO4)2 * 1mol/310 g = 0.500 mol Ca3(PO4)2 0.500 mol Ca3(PO4)2 * 3Ca(OH)2/1 Ca3(PO4)2 = 1.50 mol Ca(OH)2 1.50 mol Ca(OH)2 * 74 g/mol = 111 g b. 0.5 mol Ca3(PO4)2* 6 H2O/1 Ca3(PO4)2 = 3.0 mol H2O 3.0 mol H2O * 18 g/mol * 1 mL/1 g = 54 mL water c. 0.5 mol Ca3(PO4)2 * 6 NH3/1 Ca3(PO4)2 = 3.0 mol NH3 3.0 mol NH3 * 6.0221 x 1023 molecules/mol = 1.8 x 1024 molecules

  23. Limiting-Reactant and Yield Calculations The limiting reactant is one of the reactants in a chemical reaction. It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present. The limiting reactant limits the amount of product that can be formed.

  24. + H2 + Cl2 2HCl 4 molecules Cl2 can form 8 molecules HCl Cl2 is the limiting reactant 3 molecules of H2 remainH2 is in excess 7 molecules H2 can form 14 molecules HCl 9.3

  25. Steps Used to Determine the Limiting Reactant • Calculate the amount of product (moles or grams, as needed) formed from each reactant. • Determine which reactant is limiting. The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess. • Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the that substance that remains unreacted.

  26. How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reactant? H2 + Cl2 → 2 HCl Step 1 Calculate the moles of HCl that can form from each reactant. Step 2 Determine the limiting reactant. The limiting reactant is Cl2 because it producesless HCl than H2.

  27. You Practice: Work on Skillbuilders 8.4 & 8.5 with a partner. Turn in one piece of paper with both your names on it showing all work.

  28. How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted? Step 1 Calculate the grams of AgBr that can form from each reactant. MgBr2(aq) + 2 AgNO3 (aq) → 2 AgBr(s) + Mg(NO3)2(aq) The conversion needed is g reactant → mol reactant → mol AgBr → g AgBr

  29. How many grams of the excess reactant (AgNO3) remain unreacted? Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2. The conversion needed is g MgBr2→ mol MgBr2→ mol AgNO3→ g AgNO3 The amount of AgNO3 that remains is 100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO3

  30. Theoretical Yield The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation. Many reactions fail to give a 100% yield of product. This occurs because of side reactions and the fact that many reactions are reversible.

  31. Theoretical Yield & Percent Yield The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant. The actual yield is the amount of product finally obtained from a given amount of reactant. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100. Actual yield * 100 = Percent Yield Theor yield

  32. Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr2(aq) + 2AgNO3 (aq)→ 2AgBr(s) + Mg(NO3)2(aq) Step 1Determine the theoretical yield by calculating the grams of AgBr that can be formed. The conversion needed is g MgBr2→ mol MgBr2→ mol AgBr → g AgBr

  33. must have same units must have same units Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr2(aq) + 2AgNO3 (aq)→ 2AgBr(s) + Mg(NO3)2(aq) Step 2 Calculate the percent yield.

  34. You Practice: Work on Skillbuilder 8.6 and problem 60 on page 257.

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