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Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ

Chapter 3. Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ. Mole - Mass Relationships in Chemical Systems. 3.1 The Mole. 3.2 Determining the Formula of an Unknown Compound. 3.3 Writing and Balancing Chemical Equations.

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Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ

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  1. Chapter 3 Stoichiometry of Formulas and Equations ปริมาณสัมพันธ์ของสูตรและสมการ

  2. Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts of Reactant and Product 3.5 Fundamentals of Solution Stoichiometry

  3. The Mole mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This amount is 6.022x1023. The number is called Avogadro’s number and is abbreviated asN. One mole (1 mol) contains 6.022x1023 entities (to four significant figures)

  4. Counting objects of fixed relative mass. Figure 3.1 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S

  5. Figure 3.2 Oxygen 32.00 g One mole of common substances. Water 18.02 g CaCO3 100.09 g Copper 63.55 g

  6. Table 3.1 Summary of Mass Terminology Term Definition Unit Isotopic mass Mass of an isotope of an element amu amu Atomic mass Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance (also called atomic weight) amu Molecular (or formula) mass (also called molecular weight) Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Molar mass (M) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) g/mol (also called gram-molecular weight)

  7. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Table 3.2 Oxygen (O) Carbon (C) Hydrogen (H) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms 6(6.022 x 1023) atoms Atoms/mole of compound 6(6.022 x 1023) atoms 12(6.022 x 1023) atoms 6(16.00 amu) =96.00 amu Mass/moleculeof compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu Mass/mole of compound 96.00 g 72.06 g 12.10 g

  8. no. of grams Mass (g) = no. of moles x g 1 mol 1 mol No. of moles = mass (g) x M no. of grams 6.022x1023 entities No. of entities = no. of moles x 1 mol 1 mol No. of moles = no. of entities x 6.022x1023 entities Interconverting Moles, Mass, and Number of Chemical Entities

  9. Summary of the mass-mole-number relationships for elements. Figure 3.3 MASS(g) of element M (g/mol) AMOUNT(mol) of element Avogadro’s number (atoms/mol) ATOMS of element

  10. PROBLEM: amount(mol) of Ag 107.9 g Ag mass(g) of Ag mol Ag mass(g) of Fe mol Fe amount(mol) of Fe 55.85g Fe 6.022x1023atoms Fe mol Fe atoms of Fe Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element Sample Problem 3.1 (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8 g of Fe? PLAN: (a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M. multiply by M of Ag (107.9g/mol) = 3.69 g Ag 0.0342 mol Ag x SOLUTION: PLAN: (b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms. divide by M of Fe (55.85g/mol) SOLUTION: 95.8g Fe x = 1.72 mol Fe multiply by 6.022x1023 atoms/mol 1.72mol Fe x = 1.04x1024 atoms Fe

  11. Summary of the mass-mole-number relationships for compounds. Figure 3.3 MASS(g) of compound M (g/mol) chemical formula AMOUNT(mol) of elements in compound AMOUNT(mol) of compound Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound

  12. mass(g) of (NH4)2CO3 amount(mol) of (NH4)2CO3 number of (NH4)2CO3 formula units mol (NH4)2CO3 6.022x1023 formula units (NH4)2CO3 96.09 g (NH4)2CO3 mol (NH4)2CO3 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound Sample Problem 3.2 Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula unit are in 41.6 g of ammonium carbonate? PROBLEM: PLAN: After writing the formula for the compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6 g to mols using M and then the mols to formula units with Avogadro’s number. divide by M multiply by 6.022x1023 formula units/mol SOLUTION: The formula is (NH4)2CO3. M = (2 x 14.01 g/mol N)+(8 x 1.008 g/mol H) +(12.01 g/mol C)+(3 x 16.00 g/mol O) = 96.09 g/mol x 41.6 g (NH4)2CO3 x 2.61x1023 formula units (NH4)2CO3 =

  13. Mass % of element X = moles of X in formula x molar mass of X (amu) x 100 molecular (or formula) mass of compound (amu) Mass percent from the chemical formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) x 100 molecular (or formula) mass of compound (amu)

  14. PROBLEM: Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. Calculating the Mass Percents and Masses of Elements in a Sample of Compound Sample Problem 3.3 (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55g of glucose? amount(mol) of element X in 1mol compound PLAN: We have to find the total mass of glucose and the masses of the constituent elements in order to relate them. multiply by M (g/mol) of X mass(g) of X in 1mol of compound SOLUTION: divide by mass(g) of 1mol of compound Per mole glucose there are 6 moles of C 12 moles H 6 moles O (a) mass fraction of X multiply by 100 mass % X in compound

  15. 1.008 g H 12.01 g C mol H mol C 16.00 g O mol O 96.00 g O 72.06 g C 180.16 g glucose 180.16 g glucose 12.096 g H 180.16 g glucose Calculating the Mass Percents and Masses of Elements in a Sample of Compound Sample Problem 3.3 continued 12 mol H x = 12.096 g H 6 mol C x = 72.06 g C M = 180.16 g/mol 6 mol O x = 96.00 g O (b) mass percent of C = = 0.3999 x 100 = 39.99 mass %C mass percent of H = = 0.06714 x 100 = 6.714 mass %H = 0.5329 mass percent of O = x 100 = 53.29 mass %O

  16. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the empirical formula.

  17. mol Na mol Cl mol O 35.45 g Cl 22.99 g Na 16.00 g O Determining the Empirical Formula from Masses of Elements Sample Problem 3.4 PROBLEM: Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? PLAN: Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). SOLUTION: 2.82 g Na x = 0.123 mol Na mass(g) of each element divide by M(g/mol) 4.35 g Cl x = 0.123 mol Cl amount(mol) of each element use # of moles as subscripts 7.83 g O x = 0.489 mol O preliminary formula NaClO4 Na1 Cl1 O3.98 Na1 Cl1 O3.98 change to integer subscripts NaClO4 is sodium perchlorate. empirical formula

  18. Determining a Molecular Formula from Elemental Analysis and Molar Mass Sample Problem 3.5 PROBLEM: During physical activity. lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. PLAN: assume 100g lactic acid and find the mass of each element divide each mass by mol mass(M) amount(mol) of each element molecular formula use # mols as subscripts preliminary formula divide mol mass by mass of empirical formula to get a multiplier convert to integer subscripts empirical formula

  19. mol H mol O mol C 16.00 g O 1.008 g H 12.01g C 3.33 3.33 3.33 90.08 g molar mass of lactate 30.03 g mass of CH2O Determining a Molecular Formula from Elemental Analysis and Molar Mass Sample Problem 3.5 continued SOLUTION: Assuming there are 100 g of lactic acid, the constituents are 40.0 g C x 6.71 g H x 53.3 g O x = 3.33 mol C = 6.66 mol H = 3.33 mol O C3.33 H6.66 O3.33 CH2O empirical formula C3H6O3 is the molecular formula 3

  20. m 2 m 2 CnHm + (n+ ) O2 = n CO(g) + H2O(g) Figure 3.4 Combustion train for the determination of the chemical composition of organic compounds.

  21. PROBLEM: Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO2 absorber after combustion = 85.35g mass of CO2 absorber before combustion = 83.85g mass of H2O absorber after combustion = 37.96g mass of H2O absorber before combustion = 37.55g What is the molecular formula of vitamin C? Determining a Molecular Formula from Combustion Analysis Sample Problem 3.6 difference (after-before) = mass of oxidized element PLAN: find the mass of each element in its combustion product preliminary formula empirical formula molecular formula find the mols

  22. 85.35 g - 83.85 g = 1.50 g CO2 37.96 g - 37.55 g = 0.41 g H2O 2.016 g H2O 12.01 g CO2 18.02 g H2O 44.01 g CO2 12.01 g C 1.008 g H 16.00 g O 0.409 g C 0.046 g H 0.545 g O 176.12 g/mol 88.06 g Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis continued SOLUTION: 1.50 g CO 2 x There are 12.01 g C per mol CO2 = 0.409 g C There are 2.016 g H per mol H2O 0.41 g H2O x = 0.046 g H O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 = 0.0341 mol C = 0.0456 mol H = 0.0341 mol O C1H1.3O1 C3H4O3 = 2.000 Molecular formular = C6H8O6

  23. Table 3.3 Some Compounds with Empirical Formula CH2O (Composition by Mass: 40.0% C, 6.71% H, 53.3% O) Molecular Formula M (g/mol) Whole-Number Multiple Name Use or Function 30.03 1 CH2O formaldehyde disinfectant; biological preservative 60.05 2 C2H4O2 acetic acid acetate polymers; vinegar (5%soln) 3 90.09 C3H6O3 lactic acid sour milk; forms in exercising muscle 4 120.10 C4H8O4 erythrose part of sugar metabolism 5 150.13 C5H10O5 ribose component of nucleic acids and B2 6 180.16 C6H12O6 glucose major energy source of the cell CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6

  24. C4H10 C2H6O Table 3.4 Two Pairs of Constitutional Isomers Property Butane 2-Methylpropane Ethanol Dimethyl Ether M(g/mol) 58.12 58.12 46.07 46.07 Boiling Point -0.50C -11.060C 78.50C -250C Density at 200C 0.579 g/mL (gas) 0.549 g/mL (gas) 0.789 g/mL (liquid) 0.00195 g/mL (gas) Structural formulas Space-filling models

  25. The formation of HF gas on the macroscopic and molecular levels. Figure 3.6

  26. A three-level view of the chemical reaction in a flashbulb. Figure 3.7

  27. Balancing Chemical Equations translate the statement balance the atoms adjust the coefficients check the atom balance specify states of matter

  28. PLAN: SOLUTION: 2C8H18 + 25O216CO2 + 18H2O C8H18 + O2 CO2 + H2O C8H18 + O2 CO2 + H2O 2C8H18 + 25O216CO2 + 18H2O balance the atoms adjust the coefficients check the atom balance 2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g) specify states of matter Balancing Chemical Equations Sample Problem 3.7 PROBLEM: Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. translate the statement 25/2 8 9

  29. Summary of the mass-mole-number relationships in a chemical reaction. Figure 3.8 MASS(g) of compound A MASS(g) of compound B M (g/mol) of compound A M (g/mol) of compound B molar ratio from balanced equation AMOUNT(mol) of compound A AMOUNT(mol) of compound B Avogadro’s number (molecules/mol) Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound A MOLECULES (or formula units) of compound B

  30. Calculating Amounts of Reactants and Products Sample Problem 3.8 In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? PROBLEM: PLAN: write and balance equation find mols O2 find mols SO2 find mols Cu2O find g SO2 find mols O2 find kg O2

  31. 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 3mol O2 (a) 10.0mol Cu2S x 2mol Cu2S 2mol SO2 64.07g SO2 2mol Cu2S mol SO2 (b) 10.0mol Cu2S x 103g Cu2O mol Cu2O kg Cu2O 143.10g Cu2O 3mol O2 32.00g O2 kg O2 (c) 2.86kg Cu2O x 2mol Cu2O mol O2 103g O2 Calculating Amounts of Reactants and Products Sample Problem 3.8 continued SOLUTION: (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? = 15.0mol O2 (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? x = 641g SO2 (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? x = 20.0mol Cu2O x x 20.0mol Cu2O x = 0.959kg O2

  32. PROBLEM: Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process. 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 2Cu2O(s) + 2C(s) 4Cu(s) + 2CO(g) 2Cu2S(s)+3O2(g)+2C(s) 4Cu(s)+2SO2(g)+2CO(g) Writing an Overall Equation for a Reaction Sequence Sample Problem 3.9 PLAN: SOLUTION: write balanced equations for each step cancel reactants and products common to both sides of the equations sum the equations

  33. Figure 3.10 An ice cream sundae analogy for limiting reactions.

  34. molecules 1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O mass (amu) 44.09 amu C3H8 + 160.00 amu O2 132.03 amu CO2 + 72.06 amu H2O mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O total mass (g) 204.09 g 204.09 g Table 3.5 Information Contained in a Balanced Equation Viewed in Terms of Reactants C3H8(g) + 5O2(g) Products 3CO2(g) + 4H2O(g)

  35. PROBLEM: Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine. Using Molecular Depictions to Solve a Limiting-Reactant Problem Sample Problem 3.10 (a) Suppose the box shown at left represents a container of the reactant mixture before the reaction occurs (with chlorine colored green). Name the limiting reactant, and draw the container contents after the reaction is complete. (b) When the reaction is run again with 0.750 mol of Cl2 and 3.00 mol of F2, what mass of chlorine trifluoride will be prepared? PLAN: Write a balanced chemical equation. Compare the number of molecules you have to the number needed for the products. Determine the reactant that is in excess. The other reactant is the limiting reactant.

  36. Cl2(g) + 3F2(g) 2ClF3(g) Cl2 F2 3.00 mol F2 4 = 0.750 mol Cl2 1 2 mol ClF3 92.5 g ClF3 = 139 g ClF3 1 mol Cl 1 mol ClF3 Using Molecular Depictions to Solve a Limiting-Reactant Problem Sample Problem 3.10 continued SOLUTION: (a) You need a ratio of 2 Cl and 6 F for the reaction. You have 6 Cl and 12 F. 6 Cl would require 18 F. 12 F need only 4 Cl (2 Cl2 molecules). There isn’t enough F, therefore it must be the limiting reactant. You will make 4 ClF2 molecules (4 Cl, 12 F) and have 2 Cl2 molecules left over. (b) We know the molar ratio of F2/Cl2 should be 3/1. Since we find that the ratio is 4/1, that means F2 is in excess and Cl2 is the limiting reactant. 0.750 mol Cl2 x x

  37. Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant Sample Problem 3.11 PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed? PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given. In this case one of the reactants is in molar excess and the other will limit the extent of the reaction. mass of N2H4 mass of N2O4 limiting mol N2 divide by M multiply by M mol of N2H4 mol of N2O4 g N2 molar ratio mol of N2 mol of N2

  38. N2H4(l) + N2O4(l) N2(g) + H2O(l) mol N2H4 32.05g N2H4 3 mol N2 28.02g N2 2mol N2H4 mol N2 mol N2O4 92.02g N2O4 3 mol N2 mol N2O4 Sample Problem 3.11 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant continued SOLUTION: 2 3 4 1.00x102g N2H4 x N2H4 is the limiting reactant because it produces less product, N2, than does N2O4. = 3.12 mol N2H4 3.12mol N2H4 x = 4.68mol N2 4.68mol N2 x = 131g N2 2.00x102g N2O4 x = 2.17mol N2O4 2.17mol N2O4 x = 6.51mol N2

  39. A + B C (reactants) (main product) D (side products) Figure 3.11 The effect of side reactions on yield.

  40. SiO2(s) + 3C(s) SiC(s) + 2CO(g) 103 g SiO2 mol SiO2 kg SiO2 60.09 g SiO2 40.10 g SiC kg mol SiC 103g 51.4 kg 66.73 kg Calculating Percent Yield Sample Problem 3.12 PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process? PLAN: SOLUTION: write balanced equation x 100.0 kg SiO2 x = 1664 mol SiO2 find mol reactant & product mol SiO2 = mol SiC = 1664 find g product predicted 1664 mol SiC x = 66.73 kg x actual yield/theoretical yield x 100 percent yield x 100 =77.0%

  41. PLAN: Molarity is the number of moles of solute per liter of solution. 0.715 mol glycine 1000mL 495 mL soln 1 L Calculating the Molarity of a Solution Sample Problem 3.13 PROBLEM: Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL? mol of glycine SOLUTION: divide by volume x concentration(mol/mL) glycine 103mL = 1L = 1.44 M glycine molarity(mol/L) glycine

  42. MASS (g) of compound in solution AMOUNT (mol) of compound in solution MOLECULES (or formula units) of compound in solution VOLUME (L) of solution Figure 3.12 Summary of mass-mole-number-volume relationships in solution. M (g/mol) Avogadro’s number (molecules/mol) M (g/mol)

  43. 0.460 moles 1 L 141.96 g Na2HPO4 mol Na2HPO4 Calculating Mass of Solute in a Given Volume of Solution Sample Problem 3.14 PROBLEM: A “buffered” solution maintains acidity as a reaction occurs. In living cells phosphate ions play a key buffering role, so biochemistry often study reactions in such solutions. How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? PLAN: Molarity is the number of moles of solute per liter of solution. Knowing the molarity and volume leaves us to find the # moles and then the # of grams of solute. The formula for the solute is Na2HPO4. volume of soln SOLUTION: multiply by M moles of solute 1.75 L x = 0.805 mol Na2HPO4 multiply by M 0.805 mol Na2HPO4 x grams of solute = 114 g Na2HPO4

  44. A • Weigh the solid needed. • Transfer the solid to a volumetric flask that contains about half the final volume of solvent. C Add solvent until the solution reaches its final volume. B Dissolve the solid thoroughly by swirling. Laboratory preparation of molar solutions. Figure 3.12

  45. Converting a concentrated solution to a dilute solution. Figure 3.13

  46. 0.15 mol NaCl L soln L solnconc 6 mol Preparing a Dilute Solution from a Concentrated Solution Sample Problem 3.14 PROBLEM: “Isotonic saline” is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotomic saline from a 6.0 M stock solution? PLAN: It is important to realize the number of moles of solute does not change during the dilution but the volume does. The new volume will be the sum of the two volumes, that is, the total final volume. MdilxVdil = #mol solute = MconcxVconc volume of dilute soln multiply by M of dilute solution SOLUTION: moles of NaCl in dilute soln = mol NaCl in concentrated soln 0.80 L soln x = 0.12 mol NaCl divide by M of concentrated soln 0.12 mol NaCl x = 0.020 L soln L of concentrated soln

  47. L HCl mass Mg(OH)2 divide by M divide by M mol HCl mol Mg(OH)2 mol ratio Calculating Amounts of Reactants and Products for a Reaction in Solution Sample Problem 3.15 PROBLEM: Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide? PLAN: Write a balanced equation for the reaction; find the grams of Mg(OH)2; determine the mol ratio of reactants and products; use mols to convert to molarity.

  48. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) mol Mg(OH)2 58.33g Mg(OH)2 2 mol HCl 1 mol Mg(OH)2 1L 0.10mol HCl Sample Problem 3.15 Calculating Amounts of Reactants and Products for a Reaction in Solution continued SOLUTION: 0.10g Mg(OH)2 x = 1.7x10-3 mol Mg(OH)2 = 3.4x10-3 mol HCl 1.7x10-3 mol Mg(OH)2 x 3.4x10-3 mol HCl x = 3.4x10-2 L HCl

  49. Solving Limiting-Reactant Problems for Reactions in Solution Sample Problem 3.16 PROBLEM: Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of mercury(II) sulfide form? PLAN: As usual, write a balanced chemical reaction. Since this is a problem concerning a limiting reactant, we proceed as we would for a limiting reactant problem. Find the amount of product which would be made from each reactant. Then choose the reactant that gives the lesser amount of product.

  50. Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq) L of Na2S L of Hg(NO3)2 multiply by M multiply byM 1 mol HgS mol Na2S mol Hg(NO3)2 1 mol Na2S 1 mol HgS mol ratio mol ratio 1 mol Hg(NO3)2 mol HgS mol HgS 232.7g HgS 1 mol HgS Sample Problem 3.16 Solving Limiting-Reactant Problems for Reactions in Solution continued SOLUTION: 0.050 L Hg(NO3)2 0.020 L Hg(NO3)2 x 0.010 mol/L x 0.10 mol/L x x = 5.0x10-4 mol HgS = 2.0x10-3 mol HgS Hg(NO3)2 is the limiting reagent. 5.0x10-4 mol HgS x = 0.12 g HgS

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