1. 3-1 Dr. Wolf’s CHM 101 Chapter 3 Stoichiometry

2. 3-2 Dr. Wolf’s CHM 101 Mole - Mass Relationships in Chemical Systems 3.5 Fundamentals of Solution Stoichiometry 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. 3-3 Dr. Wolf’s CHM 101 mole - the amount of a substance that contains the same number of entities as there are atoms in exactly 12g of carbon-12, i.e. the numerical value of the atom’s mass in grams This amount is 6.022x10 23. The number is called Avogadro’s number and is abbrieviated as N. One mole (1 mol) contains 6.022x10 23 entities (to four significant figures) The Mole

4. 3-4 Dr. Wolf’s CHM 101 Counting Objects of Fixed Relative Mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g So equal numbers will always have the same 7:4 ratio = 84:48 55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S These values come from the atomic mass values for Fe and S in the Periodic Table

5. 3-5 Dr. Wolf’s CHM 101 Water, H 2 O 18.02 g One Mole of Common Substances CaCO 3 100.09 g Oxygen, O 2 32.00 g Copper 63.55 g

6. 3-6 Dr. Wolf’s CHM 101 Table 3.1 Summary of Mass Terminology TermDefinitionUnit Isotopic massMass of an isotope of an elementamu Atomic mass Molecular (or formula) mass Molar mass ( M ) (also called atomic weight) (also called molecular weight) (also called gram-molecular weight) amu g/mol Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units)

7. 3-7 Dr. Wolf’s CHM 101 Calculating the Molar Mass of a Substance For monatomic elements, the molar mass is the numerical value on the periodic table expressed in g/mol For molecules, the molar mass is the sum of the molar masses of each of the atoms in the molecular formula.

8. 3-8 Dr. Wolf’s CHM 101 Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) Oxygen (O) Mass/mole of compound 6 atoms 96.00 g Carbon (C)Hydrogen (H) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/molecule of compound 6 atoms 12 atoms 6 moles of atoms 12 moles of atoms 6 moles of atoms 6(6.022 x 10 23 ) atoms 12(6.022 x 10 23 ) atoms 6(6.022 x 10 23 ) atoms 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu 72.06 g12.10 g So for glucose with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atom, the molar mass is 72.06 + 12.10 + 96.00 = 180.16 g/mol

9. 3-9 Dr. Wolf’s CHM 101 Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol No. of moles = mass (g) x no. of grams 1 mol No. of entities = no. of moles x 6.022x10 23 entities 1 mol No. of moles = no. of entities x 6.022x10 23 entities 1 mol

10. 3-10 Dr. Wolf’s CHM 101

11. 3-11 Dr. Wolf’s CHM 101 Sample Problem 3.1Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element PROBLEM: PLAN: SOLUTION: amount(mol) of Ag mass(g) of Ag (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? (a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M. (b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms. multiply by M of Ag (107.9g/mol) 0.0342mol Ag x mol Ag 107.9g Ag = 3.69g Ag PLAN: mass(g) of Fe amount(mol) of Fe atoms of Fe SOLUTION: 95.8g Fe x 55.85g Fe mol Fe = 1.72mol Fe 1.72mol Fe x 6.022x10 23 atoms Fe mol Fe = 1.04x10 24 atoms Fe divide by M of Fe (55.85g/mol) multiply by 6.022x10 23 atoms/mol

12. 3-12 Dr. Wolf’s CHM 101 Sample Problem 3.2Calculating the Moles and Number of Formula Units in a Given Mass of a Compound PROBLEM: PLAN: SOLUTION: 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 units are in 41.6g of ammonium carbonate? mass(g) of (NH 4 ) 2 CO 3 number of (NH 4 ) 2 CO 3 formula units amount(mol) of (NH 4 ) 2 CO 3 After writing the formula for the compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6g to mols using M and then the mols to formula units with Avogadro’s number. The formula is (NH 4 ) 2 CO 3. M = (2 x 14.01g/mol N)+(8 x 1.008g/mol H) +(12.01g/mol C)+(3 x 16.00g/mol O) = 96.09g/mol 41.6g (NH 4 ) 2 CO 3 x 2.61x10 23 formula units (NH 4 ) 2 CO 3 divide by M multiply by 6.022x10 23 formula units/mol mol (NH 4 ) 2 CO 3 96.09g (NH 4 ) 2 CO 3 6.022x10 23 formula units (NH 4 ) 2 CO 3 mol (NH 4 ) 2 CO 3 x=

13. 3-13 Dr. Wolf’s CHM 101 Mass % of element X = atoms of X in formula x atomic mass of X (amu) molecular (or formula) mass of compound(amu) x 100 Mass % of element X = moles of X in formula x molar mass of X (g) molecular (or formula) mass of compound (g) x 100

14. 3-14 Dr. Wolf’s CHM 101 Flow Chart of Mass Percentage Calculation moles of X in one mol of compound Mass % of XMass fraction of X mass (g) of X in one mol of compound M = (g/mol) of X Divide by mass(g) of one mol of compound Multiply by 100

15. 3-15 Dr. Wolf’s CHM 101 So for glucose with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atom, the molar mass is 72.06 + 12.10 + 96.00 = 180.16 g/mol So to calculate the mass percent of carbon in glucose, we just use the mass of carbon in one mole of glucose over the mass of one mole of glucose to get the mass fraction and then multiply by 100. Mass % C = (76.06g/mol / 180.16 g/mol) x 100 = 40.00 %

16. 3-16 Dr. Wolf’s CHM 101 Empirical and Molecular Formulas Empirical Formula - Molecular 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. The formula of the compound as it exists, it may be a multiple of the empirical formula.

17. 3-17 Dr. Wolf’s CHM 101 mass(g) of each element Sample Problem 3.4Determining the Empirical Formula from Masses of Elements PROBLEM: PLAN: SOLUTION: amount(mol) of each element empirical formula Elemental analysis of a sample of an ionic compound gave the following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What are the empirical formula and name of the compound? preliminary formula change to integer subscripts use # of moles as subscripts divide by M (g/mol) 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). 2.82g Na mol Na 22.99g Na = 0.123 mol Na 4.35g Cl mol Cl 35.45g Cl = 0.123 mol Cl 7.83g O mol O 16.00 O = 0.489 mol O Na 1 Cl 1 O 3.98 NaClO 4 Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate.

18. 3-18 Dr. Wolf’s CHM 101 assume 100g lactic acid and find the mass of each element Sample Problem 3.5Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: PLAN: amount(mol) of each element During physical activity. lactic acid (M=90.08g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental anaylsis shows that it 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. preliminary formula empirical formula divide each mass by mol mass(M) molecular formula use #mols as subscripts convert to integer subscripts divide mol mass by mass of empirical formula to get a multiplier

19. 3-19 Dr. Wolf’s CHM 101 Sample Problem 3.5Determining a Molecular Formula from Elemental Analysis and Molar Mass continued SOLUTION:Assuming there are 100.g of lactic acid, the constituents are 40.0g C6.71g H53.3g Omol C 12.01g C mol H 1.008g H mol O 16.00g O 3.33mol C6.66mol H3.33mol O C 3.33 H 6.66 O 3.33 3.33 CH 2 Oempirical formula mass of CH 2 O molar mass of lactate90.08g 30.03g 3 C 3 H 6 O 3 is the molecular formula

20. 3-20 Dr. Wolf’s CHM 101 Combustion Train for the Determination of the Chemical Composition of Organic Compounds. C n H m + (n+ ) O 2 = n CO 2 (g) + H 2 O (g) m 4 m 2 Combustion Analysis Determining Composition by Looking at amounts of Products of Combustion

21. 3-21 Dr. Wolf’s CHM 101 difference (after-before) = mass of oxidized element Sample Problem 3.6Determining a Molecular Formula from Combustion Analysis PLAN: find the mass of each element in its combustion product molecular formula 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 CO 2 absorber after combustion=85.35g mass of CO 2 absorber before combustion=83.85g mass of H 2 O absorber after combustion=37.96g mass of H 2 O absorber before combustion=37.55g What is the molecular formula of vitamin C? find the mols preliminary formula empirical formula

22. 3-22 Dr. Wolf’s CHM 101 SOLUTION: CO 2 85.35g-83.85g = 1.50gH2OH2O37.96g-37.55g = 0.41g There are 12.01g C per mol CO 2. 1.50g CO 2 12.01g C 44.01g CO 2 = 0.409g C 0.41g H 2 O 2.016g H 18.02g H 2 O = 0.046g H There are 2.016g H per mol H 2 O. O must be the difference: 1.000g - (0.409 + 0.049) = 0.545 0.409g C 12.01g C 0.046g H 1.008g H 0.545g O 16.00g O = 0.0341mol C= 0.0461mol H= 0.0341mol O C 1 H 1.3 O 1 C3H4O3C3H4O3 176.12g/mol 88.06g = 2.000 C6H8O6C6H8O6 Sample Problem 3.6Determining a Molecular Formula from Combustion Analysis continued

23. 3-23 Dr. Wolf’s CHM 101 Writing and Balancing Chemical Equations A chemical equation shows reactants going to products. In addition, it must be balanced....meaning the same number of each kind of atom must appear on both sides of the equation. Example: hydrogen and fluorine to give hydrogen fluoride

24. 3-24 Dr. Wolf’s CHM 101 H 2 + F 2 2 HF

25. 3-25 Dr. Wolf’s CHM 101 A three-level view of the chemical reaction in a flashbulb

26. 3-26 Dr. Wolf’s CHM 101 translate the statement Sample Problem 3.7 Balancing Chemical Equations PROBLEM: PLAN:SOLUTION: balance the atomsspecify states of matter Within the cylinders of a car’s engine, the hydrocarbon octane (C 8 H 18 ), 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. adjust the coefficients check the atom balance C 8 H 18 + O 2 CO 2 + H 2 O 825/29 2C 8 H 18 + 25O 2 16CO 2 + 18H 2 O 2C 8 H 18 ( l ) + 25O 2 ( g ) 16CO 2 ( g ) + 18H 2 O ( g ) The mole ratios are the same as the molecular coefficients in the balanced reaction.

27. 3-27 Dr. Wolf’s CHM 101 Calculation Amounts of Reactants and Products in a Chemical Reaction

28. 3-28 Dr. Wolf’s CHM 101 Sample Problem 3.8Calculating Amounts of Reactants and Products PROBLEM:In a lifetime, the average American uses 1750lb(794g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistage process. After an initial grinding step, the first stage 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 10.0mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86Kg of copper(I) oxide? PLAN:write and balance equation find mols O 2 find mols SO 2 find g SO 2 find mols Cu 2 O find mols O 2 find kg O 2

29. 3-29 Dr. Wolf’s CHM 101 SOLUTION: Sample Problem 3.8Calculating Amounts of Reactants and Products continued 2Cu 2 S( s ) + 3O 2 ( g ) 2Cu 2 O( s ) + 2SO 2 ( g ) 3mol O 2 2mol Cu 2 S 10.0mol Cu 2 S= 15.0mol O 2 10.0mol Cu 2 S 2mol SO 2 2mol Cu 2 S 64.07g SO 2 mol SO 2 = 641g SO 2 2.86kg Cu 2 O 10 3 g Cu 2 O kg Cu 2 O = 0.960kg O 2 kg O 2 10 3 g O 2 mol Cu 2 O 143.10g Cu 2 O = 20.0mol Cu 2 O 20.0mol Cu 2 O 3mol O 2 2mol Cu 2 O 32.00g O 2 mol O 2 (a) (b) (c)

30. 3-30 Dr. Wolf’s CHM 101 Sample Problem 3.9Calculating Amounts of Reactants and Products in a Reaction Sequence 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. PLAN: write balanced equations for each step cancel reactants and products common to both sides of the equations sum the equations SOLUTION: 2Cu 2 S( s ) + 3O 2 ( g ) 2Cu 2 O( s ) + 2SO 2 ( g ) Cu 2 O( s ) + C( s ) 2Cu( s ) + CO( g ) 2Cu 2 O( s ) + 2C( s ) 4Cu( s ) + 2CO( g ) 2Cu 2 S( s )+3O 2 ( g )+2C(s) 4Cu( s )+2SO 2 ( g )+2CO(g)

31. 3-31 Dr. Wolf’s CHM 101 Sample Problem 3.10Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant PROBLEM:A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N 2 H 4 ) and dinitrogen tetraoxide(N 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10 2 g of N 2 H 4 and 2.00x10 2 g of N 2 O 4 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. mol of N 2 divide by M molar ratio mass of N 2 H 4 mol of N 2 H 4 mass of N 2 O 4 mol of N 2 O 4 limiting mol N 2 g N 2 multiply by M

32. 3-32 Dr. Wolf’s CHM 101 An Ice Cream Sundae Analogy for Limiting Reactions

33. 3-33 Dr. Wolf’s CHM 101 Sample Problem 3.10Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant continued SOLUTION: N 2 H 4 ( l ) + N 2 O 4 ( l ) N 2 ( g ) + H 2 O( l ) 1.00x10 2 g N 2 H 4 = 3.12mol N 2 H 4 mol N 2 H 4 32.05g N 2 H 4 3.12mol N 2 H 4 = 4.68mol N 2 3 mol N 2 2mol N 2 H 4 2.00x10 2 g N 2 O 4 = 2.17mol N 2 O 4 mol N 2 O 4 92.02g N 2 O 4 2.17mol N 2 O 4 = 6.51mol N 2 3 mol N 2 mol N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O 4. 4.68mol N 2 mol N 2 28.02g N 2 = 131g N 2 243

34. 3-34 Dr. Wolf’s CHM 101 Yield Calculations Actual Yield - Amount of product actually produced in the reaction. It can be expressed in grams or moles. Theoretical Yield - Amount of product if the reaction proceeded as complete as possible determined by the limiting reagent. Again, it can be expressed in grams or moles. Percent Yield - The ratio of Actual Yield to Theoretical Yield (both expressed in the same units) times 100.

35. 3-35 Dr. Wolf’s CHM 101 Sample Problem 3.11Calculating Percent Yield 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 100.0kg of sand are processed, 51.4kg of SiC are recovered. What is the percent yield of SiC in this process? PLAN: write balanced equation find mol reactant & product find g product predicted percent yield actual yield/theoretical yield x 100 SOLUTION: SiO 2 ( s ) + 3C( s ) SiC( s ) + 2CO( g ) 100.0kg SiO 2 mol SiO 2 60.09g SiO 2 10 3 g SiO 2 kg SiO 2 = 1664 mol SiO 2 mol SiO 2 = mol SiC = 1664 1664mol SiC 40.10g SiC mol SiC kg 10 3 g = 66.73kg x100=77.0% 51.4kg 66.73kg

36. 3-36 Dr. Wolf’s CHM 101 Stoichiometry of Solutions Concentration in Terms of Molarity A solution consists of a smaller amount of a substance, the solute, dissolved in a larger amount of another substance, the solvent. The concentration of the solution is expressed as the amount of solute dissolved in a given amount of solution. The term most commonly used is Molarity (M), defined as moles of solute per liter of solution.

37. 3-37 Dr. Wolf’s CHM 101 Sample Problem 3.12Calculating the Molarity of a Solution PROBLEM:Hydrobromic acid(HBr) is a solution of hydrogen bromide gas in water. Calculate the molarity of hydrobromic acid solution if 455mL contains 1.80mol of hydrogen bromide. mol of HBr concentration(mol/mL) HBr molarity(mol/L) HBr SOLUTION: PLAN:Molarity is the number of moles of solute per liter of solution. 1.80mol HBr 455 mL soln 1000mL 1 L divide by volume 10 3 mL = 1L = 3.96M

38. 3-38 Dr. Wolf’s CHM 101 Sample Problem 3.13Calculating Mass of Solute in a Given Volume of Solution PROBLEM:How many grams of solute are in 1.75L of 0.460M sodium monohydrogen phosphate? volume of soln moles of solute grams of solute multiply by M PLAN: SOLUTION: 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 Na 2 HPO 4. 1.75L0.460moles 1 L 0.805mol Na 2 HPO 4 141.96g Na 2 HPO 4 mol Na 2 HPO 4 = 0.805mol Na 2 HPO 4 = 114g Na 2 HPO 4

39. 3-39 Dr. Wolf’s CHM 101 Converting a Concentrated Solution to a Dilute Solution

40. 3-40 Dr. Wolf’s CHM 101 Sample Problem 3.14Preparing a Dilute Solution from a Concentrated Solution PROBLEM:“Isotonic saline” is a 0.15M 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 0.80L of isotonic saline from a 6.0M 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. volume of dilute soln moles of NaCl in dilute soln = mol NaCl in concentrated soln L of concentrated soln multiply by M of dilute solution divide by M of concentrated soln M dil xV dil = #mol solute = M conc xV conc SOLUTION: 0.80L soln = 0.020L soln 0.15mol NaCl L soln 0.12mol NaCl L soln conc 6mol = 0.12mol NaCl

41. 3-41 Dr. Wolf’s CHM 101

42. 3-42 Dr. Wolf’s CHM 101 Sample Problem 3.15Calculating Amounts of Reactants and Products for a Reaction in Solution 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. mass Mg(OH) 2 divide by M mol Mg(OH) 2 mol ratio mol HCl L HCl divide by M

43. 3-43 Dr. Wolf’s CHM 101 Sample Problem 3.15Calculating Amounts of Reactants and Products for a Reaction in Solution SOLUTION: continued Mg(OH) 2 ( s ) + 2HCl( aq ) MgCl 2 ( aq ) + 2H 2 O( l ) 0.10g Mg(OH) 2 mol Mg(OH) 2 58.33g Mg(OH) 2 = 1.7x10 -3 mol Mg(OH) 2 1.7x10 -3 mol Mg(OH) 2 2 mol HCl 1 mol Mg(OH) 2 = 3.4x10 -3 mol HCl 3.4x10 -3 mol HCl 1L 0.10mol HCl = 3.4x10 -2 L HCl

44. 3-44 Dr. Wolf’s CHM 101 Sample Problem 3.16Solving Limiting-Reactant Problems for Reactions in Solution PROBLEM:Mercury and its compounds have many uses, from filling 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 in Sample Problem 3.10 and find the amount of product which would be made from each reactant. We then chose the reactant which gives the lesser amount of product.

45. 3-45 Dr. Wolf’s CHM 101 SOLUTION: Sample Problem 3.16Solving Limiting-Reactant Problems for Reactions in Solution continued L of Na 2 S mol Na 2 S mol HgS multiply by M mol ratio L of Hg(NO 3 ) 2 mol Hg(NO 3 ) 2 mol HgS multiply by M mol ratio 0.050L Hg(NO 3 ) 2 x 0.010 mol/L x 1mol HgS 1mol Hg(NO 3 ) 2 0.020L Na 2 S x 0. 10 mol/L x 1mol HgS 1mol Na 2 S = 5.0x10 -4 mol HgS= 2.0x10 -3 mol HgS Hg(NO 3 ) 2 is the limiting reagent. 5.0x10 -4 mol HgS 232.7g HgS 1 mol HgS = 0.12g HgS Hg(NO 3 ) 2 ( aq ) + Na 2 S( aq ) HgS( s ) + 2NaNO 3 ( aq )

46. 3-46 Dr. Wolf’s CHM 101 End of Chapter 3