Ch 4. Chemical Quantities and Aqueous Reactions

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) 1 mol 2 mol Stoichiometry of the reaction FIXED ratio for each reaction If the amount of one chemical is known we can calculate how much other chemicals are required or produced.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) 1 mol 2 mol 2 mol 4 mol 2 mol 4 mol 3 mol 6 mol 3 mol 6 mol 5.22 mol 10.44 mol 5.22 mol 10.44 mol

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) 6.02 mol of NH3 is used in the above reaction. How many moles of O2 is required to react with all the NH3? How many moles of H2O will be produced?

2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g) 2 mol 3 mol 1 mol 3 mol 3 mol 6.02 mol x y z u

2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g) 2 mol 3 mol 1 mol 3 mol 3 mol 6.04 g mass? 6.04 g ÷ (14.01 g/mol x 1 + 1.008 g/mol x 3) = 0.355 mol 0.355 mol x y Mass of CuO = 0.533 mol x 79.55 g/mol = 42.4 g Mass of N2 = 0.178 mol x 28.02 g/mol = 4.99 g

Other Examples: page 143 ― 144

CH4 + 2O2 → CO2 + 2H2O 0 mol 1 mol 2 mol initial: 1 mol 2 mol final: The actual amount of reactants consumed and actual amount of products produced agree with the stoichiometry. 1 mol 0 mol initial: ? mol final:

CH4 + 2O2 → CO2 + 2H2O 1 mol 0 mol 1 mol x = 0.5 mol y z (1 − 0.5) mol initial: consumed: 1 mol x = 0.5 mol final: y z (1 − 0.5) mol 0 mol = 0.5 mol = 1 mol x = 0.5 mol 1 mol CH4 requires 2 mol O2, available O2 is 1 mol: limiting reagent. Result: 1 mol O2 will be consumed completely. CH4 will have leftover: excess reagent.

The reactant of which there are fewer moles than the stoichiometry requires is the limiting reagent. The reactant of which there are more moles than the stoichiometry requires is the excess reagent. Chemical reactions always occur according to the stoichiometry, therefore the limiting reagent is consumed and the excess reagent has leftover. The amount of products is determined by the amounts of reagents that are actually consumed.

CH4 + 2O2 → CO2 + 2H2O 1 mol 0 mol 0.5 mol 1 mol (1 − 0.5) mol 0 mol excess reagent limiting reagent CH4 + 2O2 → CO2 + 2H2O 1 mol 0 mol initial: consumed: 0.5 mol 1 mol (1 − 0.5) mol final: 0 mol 0.5 mol 1 mol 0.5 : 1 : 0.5 : 1 1 : 2 : 1 : 2 =

+ + + +

2 slices of bread + 1 slice if ham → 1 sandwich 1 sandwich + 2 slices of bread excess reagent limiting reagent amount of product excess reagent leftover

NH3(g) + CuO(s) → N2(g) + Cu(s) + H2O(g) For the following reaction, if a sample containing 18.1 g of NH3 reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N2 will be formed? NH3(g) + CuO(s) → N2(g) + Cu(s) + H2O(g) How many grams of excess reagent will be leftover? If 6.63 g of N2 is actually produced, what is the percent yield?

Procedure for limiting/excess reagent calculations aA + bB → cC + dD Make sure the equation is balanced. Find the moles of each reactant: moles = mass in gram / molar mass 3) Pick up any reactant, say A, and use the stoichiometry to calculate the required amount of the other reactant B. Compare the required amount of B with the available amount of B. a) If required > available, then B is the limiting reagent and A is the excess reagent. b) If required < available, then B is the excess reagent and A is the limiting reagent. Use the amount of the limiting reagent and the stoichiometry to calculate the amount of any product and the amount of the excess reagent that has been consumed. Leftover excess reagent = available − consumed If actual yield is given percent yield = (actually yield / theoretical yield) x 100%

Problem Set 7

Review experiment 11 Review problem sets 6 & 7

Midterm Exam Time: next week during lab session Material covered: up to this point Problem sets are very helpful

Classification of Matter Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable) (Solutions) Mixtures (multiple components) Pure Substances (one component) Matter Elements Compounds

Solute + Solvent = Solution Solvent = water, aqueous solution Water can dissolve many substances

H2O O H H Chapter 4, Figure 4.8 Charge Distribution in a Water Molecule

Chapter 4, Figure 4.9 Solute and Solvent Interactions in a Sodium Chloride Solution

Solution conducts electricity well Chapter 4, Figure 4.10 Sodium Chloride Dissolving in Water Solution conducts electricity well

C12H22O11 Chapter 4, Figure 4.12 Sugar and Water Interactions

Solution does not conduct electricity Chapter 4, Figure 4.13 A Sugar Solution Solution does not conduct electricity

Solution conducts electricity, but weakly Chapter 4, Unnumbered Figure 3, Page 148 Solution conducts electricity, but weakly

Based on the electrical conductivity in aqueous solution strong electrolytes weak electrolytes electrolytes nonelectrolytes solutes

strong electrolytes: dissociate 100 % into ions weak electrolytes: only a small fraction dissociate into ions nonelectrolytes: no dissociation

Chapter 4, Figure 4.14 Electrolytic Properties of Solutions

strong acids: HCl, HNO3, H2SO4, HClO4 strong bases: NaOH, KOH salts: NaCl, K2SO4, …… strong acids: HCl, HNO3, H2SO4, HClO4 strong bases: NaOH, KOH Strong electrolytes Bases: compounds that give OH− when dissolved in water. weak acids: acetic acid: HC2H3O2 weak bases: ammonia: NH3 Weak electrolytes

Reaction of NH3 in Water

concentrations no unit

10. g of sugar is dissolved in 40. g of water. What is the mass percent of sugar in this solution?

Unit: mol/L or M

Example 4.5, page 153 25.5 g of KBr is dissolved in water and forms a solution of 1.75 L. What is the molarity of the solution? Example 4.6, page 154 How many liters of a 0.125 mol/L NaOH solution contains 0.255 mol of NaOH?

How to prepare 1.00 L of NaCl aqueous solution with a molarity of 1.00 mol/L? 1.00 mol NaCl + 1.00 L of H2O = 1.00 mol/L NaCl (aq)

Chapter 4, Figure 4.5 Preparing a 1 Molar NaCl Solution

Solution Dilution Concentrated solutions for storage, called stock solutions stock solution + water  desired solution

moles of solute before dilution = moles of solute after dilution M1V1 = M2V2 M1: molarity of concentrated solution V1: volume of concentrated solution M2: molarity of diluted solution V2: volume of diluted solution Example on page 155 A lab procedure calls for 3.00 L of a 0.500 mol/L CaCl2 solution. How should we prepare it from a 10.0 mol/L stock solution?

Chapter 4, Figure 4.6 Preparing 3.00 L of 0.500 M CaCl2 from a 10.0 M Stock Solution

Example 4.7, page 156 To what volume should you dilute 0.200 L of a 15.0 mol/L NaOH solution to obtain a 3.00 mol/L NaOH solution?

Types of reactions Precipitation reactions

complete ionic equation NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq) formula equation Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq) complete ionic equation Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq)  AgCl(s) + Na+(aq) + NO3−(aq) spectator ions Cl−(aq) + Ag+(aq)  AgCl(s) net ionic equation

Predict whether each compound is soluble or insoluble. EXAMPLE 4.9 Predicting whether an Ionic Compound Is Soluble Predict whether each compound is soluble or insoluble. (a) PbCl2 (b) CuCl2 (c)Ca(NO3)2 (d) BaSO4

BaCl2(aq)  Ba2+(aq) + 2Cl−(aq) BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2KCl(aq) BaCl2(aq)  Ba2+(aq) + 2Cl−(aq) K2SO4(aq)  2K+ (aq) + SO42− (aq) Ba2+(aq) + 2Cl−(aq) + 2K+ (aq) + SO42− (aq)  BaSO4(s) + 2Cl−(aq) + 2K+(aq) Ba2+(aq) + SO42− (aq)  BaSO4(s)

BaCl2(aq) + KNO3(aq)  BaCl2(aq) + 2KNO3(aq)  Ba(NO3)2(aq) + 2KCl(aq) BaCl2(aq)  Ba2+(aq) + 2Cl−(aq) KNO3(aq)  K+ (aq) + NO3− (aq) 2KNO3(aq)  2K+ (aq) + 2NO3− (aq) Ba2+(aq) + 2Cl−(aq) + 2K+ (aq) + 2NO3− (aq)  Ba2+(aq) + 2NO3− (aq) + 2Cl−(aq) + 2K+(aq)

Types of reactions Precipitation reactions Acid-base reactions

Acid: Substance that produces H+ ions in aqueous solution Base: Substance that produces OH− ions in aqueous solution

Chapter 4, Unnumbered Figure 1, Page 156

Chapter 4, Unnumbered Figure 2, Page 156

Try to remember them

HCl(aq)  H+(aq) + Cl−(aq) NaOH(aq)  Na+(aq) + OH−(aq) H+(aq) + OH−(aq)  H2O(l) HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) H+(aq) + Cl−(aq) +Na+(aq) +OH−(aq)  H2O(l) + Na+(aq) + Cl−(aq) H+(aq) + OH−(aq)  H2O(l) neutralization acidic basic neutral

What is the molarity of HCl(aq) or H+(aq)? HCl(aq)  H+(aq) + Cl−(aq) Chapter 4, Unnumbered Figure, Page 158

When reaction completes HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) When reaction completes nNaOH = nHCl MNaOHVNaOH = MHClVHCl prepared, known measured by buret, known unknown measured by pippet, known

MNaOHVNaOH = MHClVHCl Chapter 4, Figure 4.19 Acid–Base Titration

Read acid-base titration starting on page 171. Chapter 4, Figure 4.20 Titration End point: light pink Read acid-base titration starting on page 171. Read the online instruction next week for the titration experiment.

MNaOHVNaOH = MHClVHCl HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) The titration of a 25.00-mL sample of an HCl solution of unknown concentration requires 32.54 mL of a 0.100 mol/L NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution in mol/L? HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) MNaOHVNaOH = MHClVHCl

Types of reactions Precipitation reactions Acid-base reactions Oxidation-Reduction reactions

2Mg(s) + O2(g) → 2MgO(s) Reactions that involve electron transfer are called oxidation-reduction reactions, or redox reactions.

Oxidation number (state) A way to keep track of the electrons gained or lost 1) For atoms in its elemental form, oxidation number = 0 Na, Ar, O2, N2, O3, P4, S8 2) For monatomic ion, oxidation number = charge of the ion Na+, Ca2+, Co2+, Co3+, Cl−, O2− NaCl, Na2O, CaCl2, CaO, CoCl2, CoCl3, Co2O3, CoO

H2O O H H Chapter 4, Figure 4.8 Charge Distribution in a Water Molecule

3) In covalent compounds Remember O: −2 H: +1 F: −1 In a neutral compound, the sum of the oxidation number = 0 CO, CO2, SF6, SF4, H2S, NH3, P2O5, N2O3 In a polyatomic ion, the sum of the oxidation number = ion charge NO3−, SO42−, NH4+, Cr2O72−, MnO4−

Reactions that cause change of oxidation numbers are called redox reactions. Element loses electrons → its oxidation number increases → element is oxidized → oxidation reaction Substance that contains the oxidized element is call the reducing agent. Element gains electrons → its oxidation number decreases → element is reduced → reduction reaction Substance that contains the reduced element is call the oxidizing agent.

PbO (s) + CO (g) → Pb (s) + CO2 (g)

Cu − 2e−  Cu2+

Problem Set 8 concentrations: mass percent and molarity; precipitation reactions; redox reactions.