3 Chemical Equations Reaction Stoichiometry CH 4 H

3 Chemical Equations & Reaction Stoichiometry 化學方程式 化學平衡及化學計量 CH 4 燃燒 產生 H 2 O, CO 2

Chapter Three Goals 1. Chemical Equations 化學方程式 2. Calculations Based on Chemical Equations 3. The Limiting Reactant Concept 限量反應物概念 4. Percent Yields from Chemical Reactions 5. Sequential Reactions 連續性反應 6. Concentrations of Solutions 溶液濃度 7. Dilution of solutions 溶液的稀釋 8. Using Solutions in Chemical Reactions 2

Chemical Equations 化學方程式 Chemical Equations 1. The substances that react, called reactants on left side of reaction 反應物 2. The substances formed, called products on right side of equation 產物 3. The relative amounts of the substances involved relative amounts of each using stoichiometric coefficients係數 represent the number of molecule 3

Coefficients indicate the amount of each Compound or element present and CAN be changed to balance the equation CH 4 + 2 O 2 CO 2+ 2 H 2 O Reactants Products Subscripts indicate the number of atoms of each element present in the compound or element present and CANNOT be changed when balancing an equation Chemical equations are based on experimental observation 4

Chemical Equations Low of Conservation of Matter • Attempt to show on paper what is happening at the laboratory and molecular levels. Ball-and-stick models Chemical formula Space-filling model 5

Chemical Equations • Law of Conservation of Matter ：物質守恆定律 在任何物理與化學作用中, 物質既不會被創造也不會 被消滅, 而只是從一種狀態轉變到另一種狀態 – There is no detectable change in quantity of matter in an ordinary chemical reaction. – Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. – This law was determined by Antoine Lavoisier. the father of modern chemistry • Propane, C 3 H 8, 丙烷burns in oxygen to give carbon dioxide and water. C 3 H 8 + 5 O 2 3 CO 2+ 4 H 2 O 6

Law of Conservation of Matter reactants C 2 H 6 O + O 2 products 2 CO 2+ 3 H 2 O 3 X 2 =6 atom count 2 C, 6 H, 3 O 2 C, 6 H, 7 O

Law of Conservation of Matter 4 oxygen atoms = 2 O 2 + C 2 H 6 O + O 2 2 CO 2+ 3 H 2 O C 2 H 6 O + 3 O 2 2 CO 2+ 3 H 2 O Final atom count 2 C, 6 H, 7 O C 2 H 6 O + 3 O 2 2 C, 6 H, 7 O 2 CO 2+ 3 H 2 O

Law of Conservation of Matter Al +3 HCl Al. Cl 3+ H 2 Al +6 HCl Al. Cl 3+3 H 2 Al +6 HCl 2 Al. Cl 3+ 3 H 2 2 Al +6 HCl 2 Al. Cl 3+ 3 H 2 atom count: Al, 3 H, 3 Cl atom count: Final atom count Al, 6 H, 6 Cl 2 Al, 6 H, 6 Cl 2 Al + 6 HCl Al, 2 H, 3 Cl Al, 6 H, 3 Cl 2 Al, 6 H, 6 Cl 2 Al. Cl 3+ 3 H 2

Law of Conservation of Matter • NH 3 burns in oxygen to form NO & water NH 3 + O 2 4 NH 3 + 5 O 2 NO 2+ H 2 O 4 NO 2+ 6 H 2 O C 7 H 16 + O 2 C 7 H 16 + 11 O 2 CO 2+ H 2 O 7 CO 2+ 8 H 2 O • C 7 H 16 burns in oxygen to form carbon dioxide and water. • Balancing equations is a skill acquired only with lots of practice • work many problems 10

Law of Conservation of Matter Example 3 -1 Balancing Chemical Equation Balance the following chemical equations: (a) P 4 + Cl 2 PCl 3 (b) Rb. OH + SO 2 Rb 2 SO 3 + H 2 O (b) P 4 O 10 + Ca(OH)2 Ca 3(PO 4)2 + H 2 O (a) P 4 + 6 Cl 2 4 PCl 3 (b) 2 Rb. OH + SO 2 Rb 2 SO 3 + H 2 O (b) P 4 O 10+6 Ca(OH)2 2 Ca 3(PO 4)2+6 H 2 O Exercise 8, 10 11

Calculations Based on Chemical Equations • Can work in moles, formula units, etc. • Frequently, we work in mass or weight (grams or kg or pounds or tons). Fe 2 O 3 + 3 CO reactants 1 formula unit 3 molecules 1 mole 3 moles 159. 7 g 84. 0 g 2 Fe 2+ 3 CO 2 yields products 2 atoms 2 moles 111. 7 g 3 molecules 3 moles 132 g 12

Calculations Based on Chemical Equations Example 3 -1: How many CO molecules are required to react with 25 formula units of Fe 2 O 3? Fe 2 O 3 +3 CO 2 Fe+3 CO 2 3 CO molecules ? CO molecules=25 formula units Fe 2 O 3 x 1 Fe 2 O 3 formula unit =75 molecule of CO 13

Calculations Based on Chemical Equations Example 3 -2: How many iron atoms can be produced by the reaction of 2. 50 x 105 formula units of iron (III) oxide with excess carbon monoxide? Fe 2 O 3 +3 CO 2 Fe+3 CO 2 ? Fe atoms=2. 5 x 105 formula units Fe 2 O 3 x =5. 0 x 105 Fe atoms 2 Fe atoms 1 Fe 2 O 3 formula unit 14

Calculations Based on Chemical Equations Example 3 -3: What mass of CO is required to react with 146 g of iron (III) oxide? Fe 2 O 3 +3 CO 2 Fe+3 CO 2 ? g CO=146 g Fe 2 O 3 x 1 mol Fe 2 O 3 x 3 mol CO x 28. 0 g CO 159. 7 g Fe 2 O 3 1 mol CO =76. 8 g CO 15

Calculations Based on Chemical Equations Example 3 -4: What mass of carbon dioxide can be produced by the reaction of 0. 540 mole of iron (III) oxide with excess carbon monoxide? Fe 2 O 3 +3 CO 2 Fe+3 CO 2 ? g CO 2=0. 54 mol Fe 2 O 3 x 3 mol CO 2 x 44. 0 g CO 2 1 mol Fe 2 O 3 1 mol CO 2 =71. 3 g CO 2 16

Calculations Based on Chemical Equations • Example 3 -5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8. 65 grams? Fe 2 O 3 +3 CO 2 Fe+3 CO 2 ? g Fe 2 O 3=8. 65 g CO 2 x 1 mol CO 2 x 1 mol Fe 2 O 3 x 159. 7 g Fe 2 O 3 1 mol Fe 2 O 3 44. 0 g CO 2 3 mol CO 2 =10. 5 g Fe 2 O 3 17

Calculations Based on Chemical Equations Example 3 -6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide? Fe 2 O 3 +3 CO 2 Fe+3 CO 2 ? lb CO =125 lb Fe 2 O 3 x 454 g Fe 2 O 3 x 3 mol CO 1 lb Fe 2 O 3 159. 7 Fe 2 O 3 28 g CO 1 lb CO x x =65. 7 lb 1 mol CO 454 g CO CO YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!! 18

Calculations Based on Chemical Equations Example 3 -2, 3 -3 Number of Molecules, Number of Moles Formed 1. How many O 2 molecule react with 47 CH 4 molecules according to the preceding equation? 2. How many moles of water could be produced by the reaction of 3. 5 mol of methane with excess oxygen? CH 4 +2 O 2 CO 2+2 H 2 O ? O 2 molecules =47 CH 4 molecules x 2 O 2 molecule 1 CH 4 molecule =94 O 2 molecule ? mol H 2 O =3. 5 mol CH 4 x 2 mol H 2 O 1 mol CH 4 =7. 0 mol H 2 O Exercise 12, 14, 18 19

Example 3 -4, 3 -5, 3 -6, 3 -7 Mass of a Reactant Required, Mass of a Product Formed 1. What mass of O 2 is required to react completely with 1. 2 mol of CH 4? 2. What mass of O 2 is required to react completely with 24. 0 g of CH 4? 3. What mass of CH 4, is required to react with 96. 0 g of O 2? 4. Calculated the mass of CO 2, in grams, that can be produced by burning 6. 0 mol of CH 4 in excess O 2. CH 4 +2 O 2 CO 2+2 H 2 O ? O 2 g =1. 2 mol CH 4 x 2 mol O 2 x 32. 0 g O 2 =76. 8 g O 2 1 mol CH 4 1 mol O 2 ? O 2 mol =24 g CH 4 x 1 mol CH 4 x 2 mol O 2 x 32. 0 g O 2 =96. 0 g O 2 16. 0 g CH 4 1 mol O 2 ? CH 4 mol =96 g O 2 x 1 mol O 2 x 1 mol CH 4 x 16 g CH 4 =24. 0 g CH 4 2 mol O 2 1 mol CH 4 32 g O 2 ? CO 2 g =6. 0 mol CH 4 x 1 mol CO 2 x 44. 0 g CO 2 =264. 0 g CO 2 1 mol CH 4 1 mol CO 2 Exercise 22, 24, 26, 28 20

Calculations Based on Chemical Equations Example 3 -8 Mass of a Reactant Required Phosphorus, P 4, burns with excess oxygen to form tetraphosphorus decoxide, P 4 O 10. In this reaction, what mass of P 4 reacts with 1. 5 mol of O 2? P 4 +5 O 2 P 4 O 10 ? P 4 g =1. 5 mol O 2 x 1 mol P 4 x 124. 0 g P 4 =37. 2 g P 4 1 mol P 4 5 mol O 2 Exercise 28 21

Limiting Reactant Concept限量反應物 • Kitchen example of limiting reactant concept. 1 packet of muffin mix + 2 eggs + 1 cup of milk ® 12 muffins • How many muffins can we make with the following amounts of mix, eggs, and milk? • Mix Packets 1 2 3 4 5 6 7 Eggs Milk 1 dozen 1 gallon limiting reactant is the muffin mix 1 dozen 1 gallon 1 dozen 1 gallon limiting reactant is the dozen eggs 22

Limiting Reactant Concept CO + 2 H 2 CH 3 OH C O H H 2 molecule: excess That is: Not enough CO molecules to react with all H 2 molecule Limiting reactant (limiting reagent) 23

Example 3 -9 Limiting Reactant What mass of CO 2 could be formed by the reaction of 16. 0 g of CH 4 with 48. 0 g of O 2? CH 4 +2 O 2 CO 2+2 H 2 O CH 4 excess 1 mol 2 mol ? mol CH 4 =16. 0 g CH 4 x 1 mol CH 4 =1. 0 mol CH 4 16. 0 g CH 4 ? mol O 2 =48. 0 g O 2 x 1 mol O 2 =1. 5 mol O 2 32. 0 g O 2 ? mol CH 4 =1. 5 mol O 2 x 1 mol CH 4 =0. 75 mol CH 4 2 mol O 2 ? g CO 2 =1. 5 mol O 2 x 1. 0 mol CO 2 x 44. 0 g CO 2 2. 0 mol O 2 1. 0 mol CO 2 =33. 0 g CO 2 Exercise 28 24

Example 3 -10 Limiting Reactant What is the maximum mass of Ni(OH)2 that could be prepared by mixing two solutions that contain 25. 9 g of Ni. Cl 2 and 10. 0 g of Na. OH, respectively? Ni. Cl 2 +2 Na. OH Ni(OH)2+2 Na. Cl Ni. Cl 2 excess ? mol Ni. Cl 2 =25. 9 g Ni. Cl 2 x 1 mol Ni. Cl 2 =0. 2 mol Ni. Cl 2 129. 6 g Ni. Cl 1 mol 2 mol 2 ? mol Na. OH =10. 0 g Na. OH x 1 mol Na. OH =0. 25 mol Na. OH 40. 0 g. Na. OH ? g Ni(OH)2 =0. 25 mol Na. OH x 1. 0 mol Ni(OH)2 x 92. 7 g Ni(OH)2 2. 0 mol Na. OH 1. 0 mol Ni(OH)2 =11. 6 g Ni(OH)2 Exercise 34, 36 25

Limiting Reactant Concept • Example 3 -8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95. 6 g of carbon disulfide with 110. g of oxygen? CS 2 +3 O 2 CO 2+2 SO 2 CS 2 excess 1 mol 2 mol 1 mol CS 2 =1. 25 mol CS 2 ? mol CS 2 =95. 6 g CS 2 x 76. 2 g CS 2 ? mol O 2 =110. 0 g O 2 x 1 mol O 2 =3. 44 mol O 2 32. 0 g O 2 1 mol 3 mol ? g SO 2 =3. 44 mol O 2 x 2. 0 mol SO 2 x 64. 0 g SO 2 3. 0 mol O 2 1. 0 mol SO 2 =146. 6 g SO 2 26

Percent Yields from Reactions • Theoretical yield is calculated by assuming that the reaction goes to completion. – Determined from the limiting reactant calculation. • Actual yield is the amount of a specified pure product made in a given reaction. – In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried. • Percent yield indicates how much of the product is obtained from a reaction. % yield = Actual yield x 100% Theoretical yield 27

Percent Yields from Reactions Example 3 -11 percent Yield A 15. 6 g sample of C 6 H 6 is mixed with excess HNO 3. We isolate 18. 0 g of C 6 H 5 NO 2. What is the percent yield of C 6 H 5 NO 2 in this reaction? C 6 H 6 +HNO 3 C 6 H 5 NO 2+H 2 O 1 mol Theoretical yield ? g C 6 H 5 NO 2=15. 6 g C 6 H 6 x 123. 0 g 1 C 6 H 5 NO 2 78. 0 g C 6 H 6 =24. 6 g C 6 H 5 NO 2 % yield = 18. 0 g x 100% 24. 6 g =73. 2% Exercise 44 28

Percent Yields from Reactions Example 3 -9: A 10. 0 g sample of ethanol, C H OH, was boiled with 2 5 excess acetic acid, CH 3 COOH, to produce 14. 8 g of ethyl acetate, CH 3 COOC 2 H 5. What is the percent yield? CH 3 COOH+C 2 H 5 OH CH 3 COOC 2 H 5+H 2 O Theoretical yield ? g CH 3 COOC 2 H 5 =10. 0 g C 2 H 5 OH x 88. 0 g CH 3 COOC 2 H 5 46. 0 g C 2 H 5 OH =19. 1 g CH 3 COOC 2 H 5 Actual yield x 100% Theoretical yield 14. 8 g x 100% % yield = 19. 1 g % yield = =77. 5% 29

Sequential Reactions Example 3 -12 Sequential Reactions At high temperatures, carbon reacts with water to produce a mixture of carbon monoxide, CO and hydrogen, H 2. Carbon monoxide is separated from H 2 and then used to separate nickel from cobalt by forming a gaseous compound, nickel tetracarbonyl, Ni(CO) 4. What mass of Ni(CO)4 could be obtained from the CO produced by the reaction of 75. 0 g of carbon? Assume 100% yield. C + H 2 O 1 mol CO+H 2 1 mol Ni + 4 CO 1 mol 4 mol Ni(CO)4 1 mol ? g CO=75. 0 g C x 1 mol C X 1 mol CO =6. 25 mol CO 12. 0 g C 1 mol C ? g Ni(CO)4=6. 25 mol CO x 1 mol Ni(CO)4 X 171 g Ni(CO)4 4 mol CO 1 mol Ni(CO)4 =267. 2 g Ni(CO)4 Exercise 50 30

Example 3 -13 Sequential Reactions Phosphoricacud, H 3 PO 4, is a very important compound used to make fertilizers. It is also present in cola drink. H 3 PO 4 can be prepared in a two-step process. Reaction 1: Reaction 2: P 4 +5 O 2 1 mol 5 mol P 4 O 10 1 mol P 4 O 10+6 H 2 O 1 mol 6 mol 4 H 3 PO 4 4 mol We allow 272 g of phosphorus to react with excess oxygen, which forms tetraphosphorus decoxide, P 4 O 10, in 89. 5% yield. In the second step reaction, a 96. 8% yield of H 3 PO 4 is obtained. What mass of H 3 PO 4 is obtained? ? g P 4 O 10=272 g P 4 x 1 mol P 4 O 10 x 284 g P 4 O 10 X 89. 5% 1 mol P 4 O 10 1 mol P 4 124 g P 4 =558 g P 4 O 10 ? g H 3 PO 4= 558 g P 4 O 10 x 1 mol P 4 O 10 x 4 mol H 3 PO 4 x 98. 0 g H 3 PO 4 1 mol H 3 PO 4 284 g P 4 O 10 1 mol P 4 O 10 x 96. 8% =746 g H PO 3 4 Exercise 52, 54 31

Concentration of Solutions • Solution is a mixture of two or more substances dissolved in another. – Solute 溶質 the dissovled phase of a solution – Solvent 溶劑the dispersed medium of a solution – In aqueous solutions 水溶液, the solvent is water. • The concentration of a solution defines the amount of solute dissolved in the solvent. – The amount of sugar in sweet tea can be defined by its concentration. • One common unit of concentration is: Mass of solute x 100% % by mass of solute = Mass of solution= mass of solute + mass of solvent % by mass of solute has the symbol % w/w 32

Concentration of Solutions Example 3 -11: What mass of Na. OH is required to prepare 250. 0 g of solution that is 8. 00% w/w Na. OH? % by mass of solute = Mass of solute x 100% Mass of solution 8. 0 g Na. OH 250. 0 g solution x =20. 0 g Na. OH 100. 0 g solution Example 3 -12: Calculate the mass of 8. 00% w/w Na. OH solution that contains 32. 0 g of Na. OH. ? g solution= 32. 0 g Na. OH x 100. 0 g solution 8. 0 g Na. OH =400. 0 g solution 33

Concentration of Solutions Example 3 -13: Calculate the mass of Na. OH in 300. 0 m. L of an 8. 00% w/w Na. OH solution. Density is 1. 09 g/m. L. sol’n X 8. 0 g Na. OH ? g solution= 300. 0 ml sol’n x 1. 09 g 1 ml sol’n 100. 0 g sol’n =26. 2 g Na. OH Sol’n weight Example 3 -14: What volume of 12. 0% KOH contains 40. 0 g of KOH? The density of the solution is 1. 11 g/m. L. ? ml solution= 40. 0 g KOH x 100. 0 g sol’n X 1 m. L sol’n 12. 0 g KOH 1. 1 g sol’n =300 ml solution Sol’n weight 34

Concentration of Solutions Example 3 -14 Percent of Solute Calculated the mass of nickel(II) sulfide, Ni. SO 4, contained in 200. 0 g of a 6% solution of Ni. SO 4. 6. 0 g Ni. SO 4 ? g Ni. SO 4= 200. 0 g sol’n x 100. 0 g solution =12. 0 g Ni. SO 4 Example 3 -15 Mass of Solution A 6% Ni. SO 4 solution contained 40. 0 g Ni. SO 4. Calculate the mass of the solution. 100. 0 g soln ? g sol’n= 40. 0 g Ni. SO 4 x 6. 0 g Ni. SO =667 g soln 4 35

Concentration of Solutions Example 3 -16 Mass of Solute Calculated the mass of Ni. SO 4 present in 200. 0 ml of a 6% solution of Ni. SO 4. The density of the solution is 1. 06 g/ml at 25 o. C. ? g Ni. SO 4= 200. 0 ml sol’n x 1. 06 g sol X 6. 0 g Ni. SO 4 1. 0 ml sol 100. 0 g solution =12. 70 g Ni. SO 4 Exercise 58 Example 3 -17 Percent solute and Density What volume of a solution that is 15% iron(III) nitrate contains 30. 0 g of Fe(NO 3) 3? The density of the solution id 1. 16 g/ml at 25 o. C. 100. 0 g sol’n x 1. 16 g sol 30. 0 g Fe(NO ) x ? ml sol’n= 3 3 15. 0 g Fe(NO ) 1. 0 ml sol 3 3 =172 ml soln Exercise 60 36

Concentrations of Solutions • Second common unit of concentration: Molarity is defined as the number of moles of solute per literof solution. Molarity = M= Number of moles of solute Number of liters of solution moles L M= mmoles m. L 37

0. 010 M KMn. O 4 sol. 0. 395 g KMn. O 4 (MW 158) in 250 ml distilled H 2 O 38

Concentrations of Solutions Example 3 -15: Calculate the molarity of a solution that contains 12. 5 g of sulfuric acid in 1. 75 L of solution. ? mol H 2 SO 4 = L sol’n 12. 5 g H 2 SO 4 98. 1 g H 2 SO 4 = 0. 0728 M H 2 SO 4 1. 75 L sol’n Example 3 -16: Determine the mass of calcium nitrate required to prepare 3. 50 L of 0. 800 M Ca(NO 3)2. ? g Ca(NO 4) 2= 3. 50 L x 0. 8 mol Ca(NO 4) 2 1. 0 L sol x 164. 0 g Ca(NO 4) 2 1 mol Ca(NO 4) 2 =459. 0 g Ca(NO 4) 2 39

Concentrations of Solutions Example 3 -17: The specific gravity of concentrated HCl is 1. 185 and it is 36. 31% w/w HCl. What is its molarity? Specific gravity =1. 185 density=1. 185 g/m. L of 1185 g/L ? M HCl= 1185 g sol 1. 0 L sol x 36. 31 g HCl x 1 mol HCl 36. 46 g HCl 100 g sol =11. 80 M HCl 40

Concentration of Solutions Example 3 -18 Molarity Calculated the Molarity (M) a solution that contains 3. 65 g of HCl in 2. 00 L of solution. ? mol HCl L sol’n = 3. 65 g HCl 36. 5 g HCl 2. 00 L sol’n = 0. 05 M HCl Exercise 62 Example 3 -19 Mass of solute Calculate the mass of Ba(OH)2 required to prepare 2. 50 L of 0. 0600 M solution of barium hydroxide. 0. 06 mol Ba(OH) 2 X 171. 3 g Ba(OH) 2 ? g Ba(OH) 2= 2. 50 L x 1. 0 L sol 1 mol Ba(OH) 2 =25. 7 g Ba(OH) 2 Exercise 64 41

Concentration of Solutions Example 3 -20 Molarity A sample of commercial sulfuric acid is 96. 4% H 2 SO 4 by mass, and its specific gravity is 1. 84. Calculate the molarity of this sulfuric acid solution. Specific gravity =1. 84 density=1. 84 g/m. L of 1840 g/L ? M H 2 SO 4= 1840 g sol 1. 0 L sol x 96. 4 g H 2 SO 4 1 mol H 2 SO 4 x 98. 1 g H 2 SO 4 100 g sol =18. 1 M H 2 SO 4 Exercise 70 42

Concentrations of Solutions • One of the reasons that molarity is commonly used is because: M x L = moles solute and M x m. L = mmol solute 43

Dilution of Solutions 溶液的稀釋 • To dilute a solution, add solvent to a concentrated solution. – One method to make tea “less sweet. ” – How fountain drinks are made from syrup. • The number of moles of solute in the two solutions remains constant. • The relationship M 1 V 1 = M 2 V 2 is appropriate for dilutions, but not for chemical reactions. 44

Dilution of Solutions • Common method to dilute a solution involves the use of volumetric flask定量瓶, pipet, and suction bulb. 100 ml 0. 1 M K 2 Cr. O 4 1 L 0. 1 M K 2 Cr. O 4 0. 01 M K 2 Cr. O 4 45

Dilution of Solutions Example 3 -18: If 10. 0 m. L of 12. 0 M HCl is added to enough water to give 100. m. L of solution, what is the concentration of the solution? M 1 V 1 = M 2 V 2 12. 0 M X 10. 0 ml = M 2 x 100. 0 ml M 2 = 1. 2 M Example 3 -19: What volume of 18. 0 M sulfuric acid is required to make 2. 50 L of a 2. 40 M sulfuric acid solution? 18. 0 M x V 1 L = 2. 40 M x 2. 5 L V 1 = 2. 40 M x 2. 5 L V 1 = 0. 333 L 18. 0 M 46

Dilution of Solutions Example 3 -21 Dilution How many milliliters of 18. 0 M H 2 SO 4 are required to prepare 1. 00 L of a 0. 9 M solution of H 2 SO 4. M 1 V 1 = M 2 V 2 18. 0 M x V 1 ml = 0. 9 M x 1000. 0 ml V 1 = 50. 0 ml Example 3 -22 Amount of solute Calculate (a) the number of moles of H 2 SO 4 and (b) the number of grams of H 2 SO 4 in 500 ml of 0. 324 M H 2 SO 4 solution. moles ? mol H 2 SO 4 = 0. 5 L x 0. 324 M =0. 162 mole V ? g H 2 SO 4 =0. 162 mole x 98. 1 =15. 9 g M=

Using Solutions in Chemical Reactions • Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution. 48

Using Solutions in Chemical Reactions Example 3 -20: What volume of 0. 500 M Ba. Cl is 2 required to completely react with 4. 32 g of Na 2 SO 4? Na 2 SO 4+Ba. Cl 2 Ba. SO 4+ 2 Na. Cl ? mol Ba. Cl 2=4. 32 g Na 2 SO 4 x 1 mol Ba. Cl 2 142 g Na 2 SO 4 1 mol Na 2 SO 4 =0. 03 mol Ba. Cl 2 moles M= V moles 0. 03 mol Ba. Cl 2 V= =0. 06 L Ba. Cl 2 = 0. 5 M Ba. Cl M 2 49

Example 3 -21: (a)What volume of 0. 200 M Na. OH will react with 50. 0 m. L 0 f 0. 200 M aluminum nitrate, Al(NO 3)3? (b)What mass of Al(OH)3 precipitates in (a)? Al(NO 3)3+3 Na. OH Al(OH)3+ 3 Na. NO 3 ? mol Al(NO 3) 3 = 50 ml/1000 mlx 0. 2 =0. 01 mol Al(NO 3)3 ? mol Na. OH =0. 01 mole Al(NO 3) 3 x 3 mol Na. OH =0. 03 mol Na. OH 1 mol Al(NO 3)2 moles = 0. 03 mol Na. OH =0. 15 L Na. OH L= M 0. 2 M Na. OH ? g Al(OH)3 =0. 01 mole Al(NO 3)3 x 1 mol Al(OH)3 x 78 g Al(OH)3 1 mol Al(NO 3)2 1 mol Al(OH)2 =0. 78 g Al(OH)3 50

Using Solutions in Chemical Reactions • Titrations 滴定are a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. – Requires special lab glassware • Buret, pipet, and flasks – Must have an indicator also 51

Example 3 -23 Solution Stoichiometry Calculate the volume in liters and in milliliters of a 0. 324 M solution of surfuric acid required to react completely with 2. 792 g of Na 2 CO 3 according to the equation. H 2 SO 4+Na 2 CO 3 Na 2 SO 4+ CO 2+H 2 O ? mol Na 2 CO 3 =2. 792 g x 1. 0 mol Na 2 CO 3 =0. 026 mole 106. 0 g Na 2 CO 3 ? mol H 2 SO 4 =0. 026 mol Na 2 CO 3 x =0. 026 mole H 2 SO 4 M= moles V ? L H 2 SO 4 = 1 mol H 2 SO 4 1 mol Na 2 CO 3 0. 026 mole 0. 324 M =0. 08 L =80. 0 ml

Example 3 -24 Volume of Solution Required Find the volume in liters and in milliliters of a 0. 505 M Na. OH solution required to react 40. 0 ml of 0. 505 M H 2 SO 4 solution according to the reaction. H 2 SO 4+2 Na. OH Na 2 SO 4+ 2 H 2 O ? mol H 2 SO 4 =0. 505 g x 40. 0 x 10 -3 L =20. 2 x 10 -3 mole ? mol Na. OH =20. 2 x 10 -3 mol H 2 SO 4 x 2 mol Na. OH 1 mol H 2 SO 4 =40. 2 x 10 -3 mole Na. OH moles M= V -3 mole 40. 2 x 10 ? L Na. OH = =0. 08 L 0. 505 M =80. 0 ml

Example 3 -22: What is the molarity of a KOH solution if 38. 7 m. L of the KOH solution is required to react with 43. 2 m. L of 0. 223 M HCl? KOH+ HCl KCl + H 2 O ? mol HCl = (43. 2 ml/1000 ml) x 0. 223 M =9. 63 x 10 -3 mol HCl ? mol KOH = 9. 63 x 10 -3 mol HCl x 1 mol KOH 1 mol HCl =9. 63 x 10 -3 mol KOH moles 9. 63 x 10= M= =0. 249 M KOH -3 V 38. 7 x 10 L KOH 54

Example 3 -23: What is the molarity of a barium hydroxide solution if 44. 1 m. L of 0. 103 M HCl is required to react with 38. 3 m. L of the Ba(OH)2 solution? Ba(OH)2 + 2 HCl Ba. Cl 2+ 2 H 2 O ? mol HCl = (44. 1 ml/1000 ml) x 0. 103 M = 4. 54 x 10 -3 mol HCl 1 mol Ba(OH)2 ? mol HCl =4. 54 x 10 -3 mol HCl x 2 mol HCl = 2. 27 x 10 -3 mol Ba(OH)2 moles M= = V 2. 27 x 10 -3 mol Ba(OH)2 -2 M KOH = 5. 93 x 10 38. 3 x 10 -3 L HCl 55