Chemical Reactions: Classification and Mass Relationships Fundamentals of General, Organic and Biological Chemistry 6th Edition Chapter Six Chemical Reactions: Classification and Mass Relationships

Outline 6.1 Chemical Equations 6.2 Balancing Chemical Equations 6.3 Avogadro’s Number and the Mole 6.4 Gram–Mole Conversions 6.5 Mole Relationships and Chemical Equations 6.6 Mass Relationships and Chemical Equations 6.7 Limiting Reagent and Percent Yield 6.8 Classes of Chemical Reactions 6.9 Precipitation Reactions and Solubility Guidelines 6.10 Acids, Bases, and Neutralization Reactions 6.13 Net Ionic Equations Chapter Six

Goals 1. How are chemical reactions written? Given the identities of reactants and products, be able to write a balanced chemical equation or net ionic equation. 2. What is the mole, and why is it useful in chemistry? Be able to explain the meaning and uses of the mole and Avogadro’s number. 3. How are molar quantities and mass quantities related? Be able to convert between molar and mass quantities of an element or compound. Chapter Six

Goals Contd. 4. What are the limiting reagent, theoretical yield, and percent yield of a reaction? Be able to take the amount of product actually formed in a reaction, calculate the amount that could form theoretically, and express the results as a percent yield. 5. How are chemical reactions of ionic compounds classified? Be able to recognize precipitation, acid–base neutralization. Chapter Six

6.1 Chemical Equations Chemical equation: An expression in which symbols are used to represent a chemical reaction. Reactant: A substance that undergoes change in a chemical reaction and is written on the left side of the reaction arrow in a chemical equation. Product: A substance that is formed in a chemical reaction and is written on the right side of the reaction arrow in a chemical equation. Chapter Six

Solid=(s) Liquid=(l) Gas=(g) Aqueous solution=(aq) The numbers and kinds of atoms must be the same on both sides of the reaction arrow. Numbers in front of formulas are called coefficients, they multiply all the atoms in a formula. The symbol 2 NaHCO3 indicates two units of sodium bicarbonate, which contains 2 Na,2 H, 2 C, and 6 O. Substances involved in chemical reactions may be solids, liquids, gases, or they may be in solution. This information is added to an equation by placing the appropriate symbols after the formulas: Solid=(s) Liquid=(l) Gas=(g) Aqueous solution=(aq) Chapter Six

6.2 Balancing Chemical Equations Balancing chemical equations can be done using four basic steps: STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element. Chapter Six

A polyatomic ion appearing on both sides of an equation can be treated as a single unit. STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. Chapter Six

is balanced, but can be simplified by dividing all coefficients by 2: STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. The equation: 2 H2SO4 + 4 NaOH  2 Na2SO4 + 4 H2O is balanced, but can be simplified by dividing all coefficients by 2: H2SO4 + 2 NaOH  Na2SO4 + 2 H2O Hint: If an equation contains a pure element as a product or reactant it helps to assign that element’s coefficient last. Chapter Six

Guidelines for Balancing Equations Start balancing with the most complicated formula first. Elements, particularly H2 & O2, should be left until the end. Balance atoms that appear in only two formulas: one as a reactant & the other as a product. Leave elements that appear in three or more formulas until later. Balance as a group those polyatomic ions that appear unchanged on both sides of the arrow. Fix the formula that contains an odd # of atoms & convert it to an even #.

Learning Check: Balancing Equations AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + NaNO3(aq) Count atoms Reactants Products 1 Ag 3 Ag 3 Na 1 Na Add in coefficients by multiplying Ag & Na by 3 to get 3 of each on both sides 3AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + 3NaNO3(aq) Now check polyatomic ions 3 NO3 3 NO3 1 PO43 1 PO43 Balanced

Learning Check __C3H8(g) + __O2(g)  __CO2(g) + __H2O(ℓ) Assume 1 in front of C3H8 3C 1C  3 8H 2H  4 1C3H8(g) + __O2(g)  3CO2(g) + 4H2O(ℓ) 2O  5 =10 O = (3  2) + 4 = 10 8H H = 2  4 = 8 1C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(ℓ)

Your Turn! Balance each of the following equations. What are the coefficients in front of each compound? __ Ba(OH)2(aq) +__ Na2SO4(aq) → __ BaSO4(s) + __ NaOH(aq) 1 1 1 2 ___KClO3(s) → ___KCl(s) +___ O2(g) 2 2 3 __H3PO4(aq) + __ Ba(OH)2(aq) → __Ba3(PO4)2(s) + __H2O(ℓ) 2 3 1 6

6.3 Avogadro’s Number and the Mole Molecular weight: The sum of atomic weights of all atoms in a molecule. Formula weight: The sum of atomic weights of all atoms in one formula unit of any compound. Mole: One mole of any substance is the amount whose mass in grams (molar mass) is numerically equal to its molecular or formula weight. Avogadro’s number: The number of molecules or formula units in a mole. NA = 6.022 x 1023 Chapter Six

6.4 Gram – Mole Conversions Molar mass = Mass of 1 mole of a substance. = Mass of 6.022 x 1023 molecules of a substance. = Molecular (formula) weight of substance in grams. Molar mass serves as a conversion factor between numbers of moles and mass. If you know how many moles you have, you can calculate their mass; if you know the mass of a sample, you can calculate the number of moles. Chapter Six

The molar mass of water is 18. 0 g The molar mass of water is 18.0 g. The conversion factor between moles of water and mass of water is 18.0 g/mol and the conversion factor between mass of water and moles of water is 1 mol/18.0 g: Chapter Six

The Mole How many atoms in 1 mole of 12C ? Avogadro’s number = NA Number of atoms in exactly 12 grams of 12C atoms How many atoms in 1 mole of 12C ? Based on experimental evidence 1 mole of 12C = 6.022 × 1023 atoms = 12.011 g Avogadro’s number = NA Number of atoms, molecules or particles in one mole 1 mole of X = 6.022 × 1023 units of X 1 mole Xe = 6.022×1023 Xe atoms 1 mole NO2 = 6.022×1023 NO2 molecules

Moles of Compounds Atoms Molecules Mass of atom (from periodic table) Atomic Mass Mass of atom (from periodic table) 1 mole of atoms = gram atomic mass = 6.022×1023 atoms Molecules Molecular Mass Sum of atomic masses of all atoms in compound’s formula 1 mole of molecule X = gram molecular mass of X = 6.022 × 1023 molecules

Moles of Compounds Ionic compounds General Formula Mass Sum of atomic masses of all atoms in ionic compound’s formula 1 mole ionic compound X = gram formula mass of X = 6.022 × 1023 formula units General Molar mass (MM) Mass of 1 mole of substance (element, molecule, or ionic compound) under consideration 1 mol of X = gram molar mass of X = 6.022 × 1023 formula units

Learning Check: Using Molar Mass Ex. How many moles of iron (Fe) are in 15.34 g Fe? What do we know? 1 mol Fe = 55.85 g Fe What do we want to determine? 15.34 g Fe = ? Mol Fe Set up ratio so that what you want is on top & what you start with is on the bottom Start End = 0.2747 mole Fe

Learning Check: Using Molar Mass Ex. If we need 0.168 mole Ca3(PO4)2 for an experiment, how many grams do we need to weigh out? Calculate MM of Ca3(PO4)2 3 × mass Ca = 3 × 40.08 g = 120.24 g 2 × mass P = 2 × 30.97 g = 61.94 g 8 × mass O = 8 × 16.00 g = 128.00 g 1 mole Ca3(PO4)2 = 310.18 g Ca3(PO4)2 What do we want to determine? 0.168 g Ca3(PO4)2 = ? Mol Fe Start End

Learning Check: Using Molar Mass Set up ratio so that what you want is on the top & what you start with is on the bottom = 52.11 g Ca3(PO4)2

Your Turn! = 0.227 mol CO2 How many moles of CO2 are there in 10.0 g? Molar mass of CO2 1 × 12.01 g = 12.01 g C 2 × 16.00 g = 32.00 g O 1 mol CO2 = 44.01 g CO2 = 0.227 mol CO2

Your Turn! How many grams of platinum (Pt) are in 0.475 mole Pt? 195 g Molar mass of Pt = 195.08 g/mol = 92.7 g Pt

6.5 Mole Relationships and Chemical Equations The coefficients in a balanced chemical equation tell how many molecules, and thus how many moles, of each reactant are needed and how many molecules, and thus moles, of each product are formed. See the example below: Chapter Six

The coefficients can be put in the form of mole ratios, which act as conversion factors when setting up factor-label calculations. In the ammonia synthesis the mole ratio of H2 to N2 is 3:1, the mole ratio of H2 to NH3 is 3:2, and the mole ratio of N2 to NH3 is 1:2 leading to the following conversion factors: (3 mol H2)/(1 mol N2) (3 mol H2)/(2 mol NH3) (1 mol N2)/(2 mol NH3) Chapter Six

Reaction Stoichiometry The coefficients in a chemical reaction specify the relative amounts in moles of each of the substances involved in the reaction. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) 2 molecules of C8H18 react with 25 molecules of O2 to form 16 molecules of CO2 and 18 molecules of H2O. 2 moles of C8H18 react with 25 moles of O2 to form 16 moles of CO2 and 18 moles of H2O. 2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O

Suppose That We Burn 22.0 Moles of C8H18; How Many Moles of CO2 Form? 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) stoichiometric ratio: 2 moles C8H18 : 16 moles CO2 The combustion of 22 moles of C8H18 adds 176 moles of CO2 to the atmosphere.

Using Stoichiometric Ratios Ex. For the reaction N2 + 3 H2 → 2NH3, how many moles of N2 are used when 2.3 moles of NH3 are produced? Assembling the tools 2 moles NH3 = 1 mole N2 2.3 mole NH3 = ? moles N2 = 1.2 mol N2

Your Turn! If 0.575 mole of CO2 is produced by the combustion of propane, C3H8, how many moles of oxygen are consumed? The balanced equation is C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) 0.575 mole 2.88 mole 0.192 mole 0.958 mole 0.345 mole Assembling the tools 0.575 mole CO2 = ? moles O2 3 moles CO2 = 5 mole O2 = 0.958 mol O2

6.6 Mass Relationships and Chemical Equations Most common stoichiometric conversions that chemists use involve converting mass of one substance to mass of another. Use molar mass A to convert grams A to moles A Use chemical equations to relate moles A to moles B Use molar mass B to convert to moles B to grams B

Mass-to-Mass Conversions Most common stoichiometric conversions that chemists use involve converting mass of one substance to mass of another. Use molar mass A to convert grams A to moles A Use chemical equations to relate moles A to moles B Use molar mass B to convert to moles B to grams B

Ex. What mass of O2 will react with 96 Ex. What mass of O2 will react with 96.1 g of propane (C3H8) gas, to form gaseous carbon dioxide & water? Strategy 1. Write the balanced equation C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) 2. Assemble the tools 96.1 g C3H8  moles C3H8  moles O2  g O2 1 mol C3H8 = 44.1 g C3H8 1 mol O2 = 32.00 g O2 1 mol C3H8 = 5 mol O2

Ex. What mass of O2 will react with 96 Ex. What mass of O2 will react with 96.1 g of propane in a complete combustion? C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) 3. Assemble conversions so units cancel correctly = 349 g of O2 are needed

Your Turn! 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(ℓ) How many grams of Al2O3 are produced when 41.5 g Al react? 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(ℓ) 78.4 g 157 g 314 g 22.0 g 11.0 g = 78.4 g Al2O3

6.7 Limiting Reagent and Percent Yield The amount of product actually formed in a chemical reaction is somewhat less than the amount predicted by theory. Unwanted side reactions and loss of product during handling prevent one from obtaining a perfect conversion of all the reactants to desired products. The amount of product actually obtained in a chemical reaction is usually expressed as a percent yield. Chapter Six

Percent yield is defined as: (Actual yield ÷ Theoretical yield) x 100% The actual yield is found by weighing the product obtained. The theoretical yield is found by a mass-to-mass calculation. When running a chemical reaction without the exact amounts of reagents to allow all of them to react completely, the reactant that is exhausted first is called the limiting reagent. Chapter Six

Apply These Concepts to a Chemical Reaction Balanced equation for the combustion of methane: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) Our balanced equation for the combustion of methane implies that every one molecule of CH4 reacts with two molecules of O2. Insert first image of molecules on Pg. 146

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant? CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) First we calculate the number of CO2 molecules that can be made from 5 CH4 molecules.

Combustion of Methane Then we calculate the number of CO2 molecules that can be made from 8 O2 molecules. We have enough CH4 to make 5 CO2 molecules and 4 CO2 molecules. Therefore, O2 is the limiting reactant, and 4 CO2 molecules is the theoretical yield. CH4 is in excess.

Ex. Limiting Reactant Calculation 2 H2 + I2 HI How many grams of HI can be formed from 2.00 g H2 and 2.00 g of I2? 2.00 g 2.00 g ? 1 mole H2 2 mole HI 127.9124g HI 2.00 g H2 -------------- ------------- ------------------- = 254 g HI 2.0158 g 1 mole H2 1 mole HI 1 mole I2 2 moles HI 127.9124g HI 2.00g I2 ------------- --------------- ----------------- = 2.02 g HI 253.810 g 1 mole I2 1 mole HI Limiting Reactant is always the smallest value! I2 is the limiting reactant and H2 is in XS.

Ex. Limiting Reactant Calculation 5 3 4 C3H8 + O2 CO2 + H2O How many grams of CO2 can be formed from 1.44 g C3H8 and 2.65 g of O2? 1.44 g 2.65g ? 1 mole C3H8 3 mole CO2 44.009 g CO2 4.31 g CO2 1.44 g C3H8 ---------------- ---------------- ------------------- = 44.0962 g 1 mole C3H8 1 mole CO2 1 mole O2 3 moles CO2 44.009 g CO2 2.19 g CO2 2.65g O2 ------------- ----------------- ------------------- = 31.998 g 5 mole O2 1 mole CO2 O2 is the limiting reactant and C3H8 is in XS.

Ex. Limiting Reactant Calculation How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to: 4 NH3 + 5 O2  4 NO + 6 H2O Solution: Step 1 mass NH3  mole NH3  mole O2  mass O2 Assembling the tools 1 mol NH3 = 17.03 g 1 mol O2 = 32.00 g 4 mol NH3  5 mol O2 Only have 40.0 g O2, O2 limiting reactant = 70.5 g O2 needed

Ex. Limiting Reactant Calculation How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to: 4 NH3 + 5 O2  4 NO + 6 H2O Solution: Step 2 mass O2  mole O2  mole NO  mass NO Assembling the tools 1 mol O2 = 32.00 g 1 mol NO = 30.01 g 5 mol O2  4 mol NO Can only form 30.0 g NO. = 30.0 g NO formed

Ex. Percentage Yield Calculation When 18.1 g NH3 and 90.4 g CuO are reacted, the theoretical yield is 72.2 g Cu. The actual yield is 58.3 g Cu. What is the percent yield? 2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g) = 80.7%

Problem: learning check When 6.40 g of CH3OH was mixed with 10.2 g of O2 and ignited, 6.12 g of CO2 was obtained. What was the percentage yield of CO2? 2CH3OH + 3O2  2CO2 + 4H2O MM(g/mol) (32.04) (32.00) (44.01) (18.02) 6.12% 8.79% 100% 142% 69.6%

6.8 Classes of Chemical Reactions When learning about chemical reactions it is helpful to group the reactions of ionic compounds into three general classes: precipitation reactions, acid–base neutralization reactions, and oxidation–reduction reactions. Precipitation reactions are processes in which an insoluble solid called a precipitate forms when reactants are combined in aqueous solution. Chapter Six

Acid–base neutralization reactions are processes in which H+ ions from an acid react with OH- ions from a base to yield water. An ionic compound called a salt is also produced. The “salt” produced need not be common table salt. Any ionic compound produced in an acid–base reaction is called a salt. Oxidation–reduction reactions, or redox reactions, are processes in which one or more electrons are transferred between reaction partners (atoms, molecules, or ions). As a result of this transfer, the charges on atoms in the various reactants change. Chapter Six

6.9 Precipitation Reactions and Solubility Guidelines Reaction of aqueous Pb(NO3)2 with aqueous KI gives a yellow precipitate of PbI2. To predict whether a precipitation reaction will occur on mixing aqueous solutions of two ionic compounds, you must know the solubility of the potential products. Chapter Six

If a potential product does not contain at least one of the ions listed below, it is probably not soluble and will precipitate from solution when formed. Chapter Six

6.10 Acids, Bases, and Neutralization Reactions When acids and bases are mixed together in correct proportion acidic and basic properties disappear. A neutralization reaction produces water and a salt. HA(aq) + MOH(aq)  H2O(l) + MA(aq) acid + base  water + salt The reaction of hydrochloric acid with potassium hydroxide to produce potassium chloride is an example: HCl(aq) + KOH(aq)  H2O(l) + KCl(aq) Chapter Six

Acid–Base Reactions Acid: Substance that produces H+ HCl(aq) H+(aq) + Cl–(aq) Some acids—called polyprotic acids These acids contain more than one ionizable proton and release them sequentially. For example, sulfuric acid, H2SO4 is a diprotic acid. Base: Substance that produces OH ions in aqueous solution NaOH(aq) Na+(aq) + OH–(aq)

2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l) Acid–Base Reactions Also called neutralization reactions because the acid and base neutralize each other’s properties 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l) The net ionic equation for an acid–base reaction is H+(aq) + OH(aq)  H2O(l)

Some Common Acids and Bases

Predict the Product of the Reactions 1. HCl(aq) + Ba(OH)2(aq) ® 2. H2SO4(aq) + LiOH(aq) ® 2 HCl(aq) + Ba(OH)2(aq) ® 2 H2O(l) + BaCl2(aq) H2SO4(aq) + 2 LiOH(aq) ® 2 H2O(l) + Li2SO4(aq)

Reactions that Release CO2 b) Acid with Bicarbonate (HCO3–) NaHCO3(aq) + HI(aq)  NaI(aq) + H2O + CO2(g) Acid with Carbonate (CO32–) CaCO3(aq) + 2HCl(aq)  CaCl2(aq) + H2O + CO2(g) Fig 5.22 and 5.23

6.13 Net Ionic Equations Ionic equation: An equation in which ions are explicitly shown. Spectator ion: An ion that appears unchanged on both sides of a reaction arrow. Net ionic equation: An equation that does not include spectator ions. Chapter Six

The Solubility of Ionic Compounds When an ionic compound dissolves in water, the resulting solution contains not the intact ionic compound itself, but its component ions dissolved in water. However, not all ionic compounds dissolve in water. If we add AgCl to water, for example, it remains solid and appears as a white powder at the bottom of the water. In general, a compound is termed soluble if it dissolves in water and insoluble if it does not.

Solubility of Salts If we mix solid AgNO3 with water, it dissolves and forms a strong electrolyte solution. Silver chloride, on the other hand, is almost completely insoluble. If we mix solid AgCl with water, virtually all of it remains as a solid within the liquid water.

Solubility Rules

Precipitation Reactions Precipitation reactions are reactions in which a solid forms when we mix two solutions. Reactions between aqueous solutions of ionic compounds produce an ionic compound that is insoluble in water. The insoluble product is called a precipitate.

Precipitation of Lead(II) Iodide

No Precipitation Means No Reaction Precipitation reactions do not always occur when two aqueous solutions are mixed. Combine solutions of KI and NaCl and nothing happens. KI(aq) + NaCl(aq)  No Reaction

Predicting Precipitation Reactions

Representing Aqueous Reactions An equation showing the complete neutral formulas for each compound in the aqueous reaction as if they existed as molecules is called a molecular equation. 2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s) In actual solutions of soluble ionic compounds, dissolved substances are present as ions. Equations that describe the material’s structure when dissolved are called ionic equations. 2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) ® 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)

2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) ® 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s) Net Ionic Equation An ionic equation in which the spectator ions are removed is called a net ionic equation. 2 OH−(aq) + Mg2+(aq) ® Mg(OH)2(s)

Write the ionic and net ionic equation for each of the following: Examples Write the ionic and net ionic equation for each of the following: 1. K2SO4(aq) + 2 AgNO3(aq) ® 2 KNO3(aq) + Ag2SO4(s) 2 K+(aq) + SO42−(aq) + 2 Ag+(aq) + 2 NO3−(aq) ® 2 K+(aq) + 2 NO3−(aq) + Ag2SO4(s) 2 Ag+(aq) + SO42−(aq) ® Ag2SO4(s) 2. Na2CO3(aq) + 2 HCl(aq) ® 2 NaCl(aq) + CO2(g) + H2O(l) 2 Na+(aq) + CO32−(aq) + 2 H+(aq) + 2 Cl−(aq) ® 2 Na+(aq) + 2 Cl−(aq) + CO2(g) + H2O(l) CO32−(aq) + 2 H+(aq) ® CO2(g) + H2O(l)

Learning Check: Convert Molecular to Ionic Equations: Write the correct ionic equation for each: Pb(NO3)2(aq) + 2NH4IO3(aq) → Pb(IO3)2(s) + 2NH4NO3(aq) 2NaCl (aq) + Hg2(NO3)2 (aq) → 2NaNO3 (aq) + Hg2Cl2 (s) Pb2+(aq) + 2NO3–(aq) + 2NH4+(aq) + 2IO3–(aq) → Pb(IO3)2(s) + 2NH4+(aq) + 2NO3–(aq) Note that Hg22+ is a polyatomic ion. 2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3–(aq) → 2Na+(aq) + 2NO3–(aq) + Hg2Cl2(s)

Your Turn Consider the following reaction : Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s) Which is the correct ionic equation? 2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl22–(aq) → 2Na+(aq) + 2Cl–(aq) + BaSO4(s) 2Na+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) → 2Na+(aq) + 2Cl–(aq) + BaSO4(s) 2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl22–(aq) → 2Na+(aq) + 2Cl–(aq) + Ba2+(s) + SO42–(s) Ba2+(aq) + SO42–(aq) → BaSO4(s) Ba2+(aq) + SO42–(aq) → Ba2+(s) + SO42–(s)

Learning Check: Convert Ionic Equation to Net Ionic Equation Write the correct net ionic equation for each. Pb2+(aq) + 2NO3–(aq) + 2K+(aq) + 2IO3–(aq) →Pb(IO3)2(s) + 2K+(aq) + 2NO3–(aq) 2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3–(aq) → 2Na+(aq) + 2NO3–(aq) + Hg2Cl2(s) Pb2+(aq) + 2IO3–(aq) → Pb(IO3)2(s) Note that Hg22+ is a polyatomic ion. 2Cl–(aq) + Hg22+(aq) → Hg2Cl2(s)