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  1. Calculations FromChemical EquationsChapter 9 Hein and Arena Eugene PasserChemistry DepartmentBronx Community College © John Wiley and Sons, Inc. Version 1.1

  2. A Short Review

  3. The molar mass of an element is its • atomic mass in grams/mol. • It contains 6.022 x 1023 atoms • (Avogadro’s number) of the element. • The molar mass of a compound is the • sum of the atomic masses of all its atoms of each element.

  4. Calculate the molar mass of C2H6O. 2 C = 2(12.01 g) = 24.02 g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = 16.00 g 46.08 g/mol

  5. 2 2  Al + Fe2O3  Fe + Al2O3 2 mol 1 mol 2 mol 1 mol • The chemical equation must be balanced by using smallest whole number coefficients. • For calculations of mole-mass-volume relationships in a chemical equation. • The coefficient in front of a formula represents the number of moles, atoms, ions or molecules, of the reactant or product. The equation is balanced.

  6. Introduction to Stoichiometry:The Mole Ratio Method

  7. Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and or products. • Mole Ratio: A ratio between the moles of any two substances involved in a chemical reaction (Conversion Factor). • The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

  8. Examples

  9. 1 mol 3 mol 2 mol N2 + 3H2 2NH3

  10. 1 mol 3 mol 2 mol N2 + 3H2 2NH3

  11. The mole ratio is used to convert moles of one substance to moles of another substance in a stoichiometry problem. mol. H2 mol. N2 mole ratio • The mole ratio is used to solve every type of stoichiometry problem.

  12. The Mole Ratio Method: Step by Step Convert the quantity of starting substance to moles (if it is not already in moles) Convert the moles of starting substance to moles of desired substance. Convert the moles of desired substance to the units specified (i.e. grams, molecules, atoms, etc.) in the problem.

  13. Mole-Mole Calculations

  14. Mole Ratio 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 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 Moles starting substance: 10.0 mol H2SO4 Step 2 The conversion sequence is …moles H2SO4 moles H3PO4

  15. Calculate the number of moles of sulfuric acid (H2SO4) that react when 10 moles of Ca5(PO4)3F react. Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4 Mole Ratio 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 The starting substance is 10.0 mol Ca5(PO4)3F Step 2 The conversion sequence is …moles Ca5(PO4)3F moles H2SO4

  16. 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) The conversion sequence is … moles NaCl moles PbCl2

  17. Mole-Mass Calculations

  18. Examples

  19. The conversion sequence is … grams H3PO4 moles H3PO4  moles H2SO4 Mole Ratio Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4 Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4

  20. Mole Ratio Calculate the number of grams of H2 required to form 12.0 moles of NH3. N2 + 3H2 2NH3 The conversion sequence is … moles NH3 moles H2  grams H2 36.4 g H2

  21. Mass-Mass Calculations

  22. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2. N2 + 3H2 2NH3 grams H2 moles H2  moles NH3  grams NH3 628 g NH3

  23. Limiting-Reactant and Yield Calculations

  24. Limiting Reactant

  25. A limiting reactant is a reactant in a chemical reaction with insufficient mass to react with all the mass of the other reactant(s) present. The limiting reactant limits the amount of product that can be formed.

  26. How many bicycles can be assembled from the parts shown? From eight wheels four bikes can be constructed. From three pedal assemblies three bikes can be constructed. From four frames four bikes can be constructed. The limiting part is the number of pedal assemblies. 9.2

  27. Steps Used to Determine the Limiting Reactant

  28. Convert moles of each reactant to moles of the other reactant (This is what you NEED). • Determine limiting reactant by: • HAVE – NEED > 0 ; NOT LIMITING • HAVE – NEED < 0 ; LIMITING • Excess reactant is the difference between: • HAVE – NEED of NON- LIMITING reactant. • Calculate the moles of each reactant (This is what you HAVE).

  29. Examples

  30. How many grams of H2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl  MgCl2 + H2 Step 1: Calculate moles of reactants you have. Step 2: Calculate moles of reactants you need. Step 3: Determine limiting reactant. Step 4: Calculate desired quantity based on limiting reactant.

  31. How many grams of H2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl  MgCl2 + H2 Have 0.31 0.41 Need Moles have (Step 1): x mol Mg = 7.5g Mg x 1mol/24.31g Mg = 0.31mol. Mg x mol HCl = 15.0g HCl x 1mol/36.45g HCL = 0.41 mol HCl

  32. How many grams of H2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl  MgCl2 + H2 Have 0.31 0.41 Need 0.21 0.62 Moles need (Step 2): x mol Mg = 0.41mol HCl x 1m Mg/2 m HCl = 0.21mol. Mg x mol HCl = 0.31 mol Mg x 2 m HCl/1m Mg = 0.62 mol HCl

  33. How many grams of H2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl  MgCl2 + H2 Have 0.31 0.41 Need 0.21 0.62 Limiting reactant (Step 3): The limiting reactant is HCl, since we have less than we need: have - need < 0; 0.41 - 0.62 = - 0.21

  34. How many grams of H2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl  MgCl2 + H2 The grams of H2 is calculated from the mass of the limiting reactant HCl. g H2  15.0g HCl x 1mol. HCl x 1mol. H2 x 2.02g H2 = 36.45g HCl 2 mo HCl 1 mol H2 0.42g H2

  35. Suppose 28.82 grams of carbon and 55.86 grams of steam react. How many grams of hydrogen will be produced? Assume 100% yield. C (s) + H2O (g)  CO (g) + H2 (g)

  36. C (s) + H2O (g)  CO (g) + H2 (g) Moles have 2.40 3.10 Moles need 3.10 2.40 0.70 moles “xs” H2O Therefore, the carbon is the limiting reactant, since we have less carbon than we need to completely react with all the water we have. The mass of hydrogen is calculated using the limiting reactant; 28.82 g C x 1 mol C x 1 mol H2 x 2.02 g H2 = 4.847 g H2 12.01 g C 1 mol C 1 mol H2

  37. Reaction Yield

  38. The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

  39. Many reactions fail to give a 100% yield of product. This occurs because of: side reactions purification steps reversible reactions

  40. 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 “actuallyobtained” from a given amount of reactant.

  41. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.

  42. 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 sequence is … g MgBr2→ mol MgBr2→ mol AgBr → g AgBr

  43. 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.

  44. The End

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