Ch 4. Chemical Quantities and Aqueous Reactions

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

1 mol 2 mol 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 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

2 mol 6.02 mol 2

3 mol x 3 x 9.03 mol

2 6.02 mol

1 y y 3.01 mol

2 6.02 mol

3 z z 9.03 mol

2 6.02 mol

3 u u 9.03 mol

2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)

2 mol 3 mol 1 mol 3 mol 3 mol

6.04 g

x y

x

mol 0.355

3

2 mol 0.533x

y

mol 0.355

1

2 mol 0.178 y

6.04 g ÷ (14.01 g/mol x 1 + 1.008 g/mol x 3) = 0.355 mol

0.355 mol

mass? mass?

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 0 mol1 mol 2 molinitial:

0 mol 0 mol 1 mol 2 molfinal:

1 mol 1 mol 0 mol 0 molinitial:

? mol ? mol ? mol ? molfinal:

The actual amount of reactants consumed and actual amountof products produced agree with the stoichiometry.

CH4 + 2O2 → CO2 + 2H2O

1 mol 1 mol 0 mol 0 molinitial:

1 mol CH4 requires 2 mol O2, available O2 is 1 mol: limiting reagent.

(1 − 0.5) mol 0 mol = 0.5 mol = 1 mol

Result: 1 mol O2 will be consumed completely.

= 0.5 molconsumed: 1 molx

mol 1

x

2

1 x = 0.5 mol

final: y z

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 thestoichiometry requires is the excess reagent.

Chemical reactions always occur according to thestoichiometry, therefore the limiting reagent is consumedand the excess reagent has leftover. The amount of productsis determined by the amounts of reagents that are actually consumed.

CH4 + 2O2 → CO2 + 2H2O

limiting reagentexcess reagent

1 mol 1 mol 0 mol 0 molinitial:

consumed: 0.5 mol 1 mol

(1 − 0.5) molfinal: 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

4 slices of bread + 1 slice if ham →

excess reagent excess reagentleftover

limiting reagent amount of product

1 sandwich + 2 slices of bread

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?

How many grams of excess reagent will be leftover?

If 6.63 g of N2 is actually produced, what is the percent yield?

% 100 yieldltheoretica

yieldactual eldpercent yi

NH3(g) + CuO(s) → N2(g) + Cu(s) + H2O(g)

1) Make sure the equation is balanced.2) Find the moles of each reactant:

moles = mass in gram / molar mass3) Pick up any reactant, say A, and use the stoichiometry to

calculate the required amount of the other reactant B.4) Compare the required amount of B with the available

amount of B.a) If required > available, then B is the limiting reagent and Ais the excess reagent.b) If required < available, then B is the excess reagent and Ais the limiting reagent.

5) Use the amount of the limiting reagent and the stoichiometryto calculate the amount of any product and the amount of theexcess reagent that has been consumed.

6) Leftover excess reagent = available − consumed7) If actual yield is given

percent yield = (actually yield / theoretical yield) x 100%

Procedure for limiting/excess reagent calculationsaA + bB → cC + dD

Problem Set 7

Review problem sets 6 & 7

Review experiment 11

Midterm Exam

Time: next week during lab session

Material covered: up to this point

Problem sets are very helpful

Classification of Matter

Matter

Elements

Compounds

Mixtures(multiple components)

Pure Substances(one component)

Homogeneous(visibly indistinguishable)

Heterogeneous (visibly distinguishable)

(Solutions)

Solute + Solvent = Solution

Solvent = water, aqueous solution

Water can dissolve many substances

O

H H

H2O

Solution conducts electricity well

C12H22O11

Solution does not conduct electricity

Solution conducts electricity, but weakly

electrolytes

nonelectrolytes

Based on the electrical conductivity in aqueous solution

strong electrolytes

weak electrolytessolutes

strong electrolytes: dissociate 100 % into ions

weak electrolytes: only a small fraction dissociate into ions

nonelectrolytes: no dissociation

salts: NaCl, K2SO4, ……

strong acids: HCl, HNO3, H2SO4, HClO4

strong bases: NaOH, KOH

Bases: compounds that give OH− when dissolved in water.

Strong electrolytes

weak acids: acetic acid: HC2H3O2

weak bases: ammonia: NH3

Weak electrolytes

Reaction of NH3 in Water

concentrations

% 100 sample wholeof mass

component of mass percent mass

% 100 solution of mass

solute of mass percent mass

no unit

10. g of sugar is dissolved in 40. g of water.

What is the mass percent of sugar in this solution?

moles of soluteMolarity (M)

liters of 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?

moles of soluteMolarity (M)

liters of solution

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)

Solution Dilution

Concentrated solutions for storage, called stock solutions

stock solution + water desired solution

solution of liters

solute of moles (M)Molarity

moles of solute before dilution = moles of solute after dilution

M1V1 = M2V2

M1: molarity of concentrated solutionV1: volume of concentrated solutionM2: molarity of diluted solutionV2: volume of diluted solution

Example on page 155

A lab procedure calls for 3.00 L of a 0.500 mol/L CaCl2solution. How should we prepare it from a 10.0 mol/L stocksolution?

Example 4.7, page 156

To what volume should you dilute 0.200 L of a 15.0 mol/LNaOH solution to obtain a 3.00 mol/L NaOH solution?

Types of reactions

Precipitation reactions

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

Cl−(aq) + Ag+(aq) AgCl(s)

net ionic equation

Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) AgCl(s) + Na+(aq) + NO3

−(aq)

spectator ions

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) + K2SO4(aq)

BaCl2(aq) Ba2+(aq) + 2Cl−(aq)

K2SO4(aq) 2K+ (aq) + SO42− (aq)

BaSO4(s) + 2KCl(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) Ba2+(aq) + 2Cl−(aq)

KNO3(aq) K+ (aq) + NO3− (aq)

BaCl2(aq) + 2KNO3(aq) Ba(NO3)2(aq) + 2KCl(aq)

Ba2+(aq) + 2Cl−(aq) + 2K+ (aq) + 2NO3− (aq)

Ba2+(aq) + 2NO3− (aq) + 2Cl−(aq) + 2K+(aq)

2KNO3(aq) 2K+ (aq) + 2NO3− (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

Try to remember them

NaOH(aq) Na+(aq) + OH−(aq)

H+(aq) + Cl−(aq) +Na+(aq) +OH−(aq) H2O(l) + Na+(aq) + Cl−(aq)

H+(aq) + OH−(aq) H2O(l)

HCl(aq) H+(aq) + Cl−(aq)

HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

H+(aq) + OH−(aq) H2O(l)

acidic basic neutral

neutralization

HCl(aq) H+(aq) + Cl−(aq)

What is the molarity ofHCl(aq) or H+(aq)?

When reaction completes

nNaOH = nHCl

MNaOHVNaOH = MHClVHCl

HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

prepared,known

measuredby buret, known

unknown measured bypippet, known

MNaOHVNaOH = MHClVHCl

Read acid-base titration starting on page 171.Read the online instruction next week for the titration experiment.

End point: light pink

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?

MNaOHVNaOH = MHClVHCl

HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

Types of reactions

Precipitation reactions

Acid-base reactions

Oxidation-Reduction reactions

Reactions that involve electron transfer are called

oxidation-reduction reactions, or redox reactions.

2Mg(s) + O2(g) → 2MgO(s)

Oxidation number (state)

1) For atoms in its elemental form, oxidation number = 0

A way to keep track of the electrons gained or lost

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

O

H H

H2O

3) In covalent compounds

Remember O: −2 H: +1 F: −1

In a neutral compound, the sum of the oxidation number = 0

In a polyatomic ion, the sum of the oxidation number = ion charge

CO, CO2, SF6, SF4, H2S, NH3, P2O5, N2O3

NO3−, SO4

2−, NH4+, Cr2O7

2−, MnO4−

Reactions that cause change of oxidation numbersare called redox reactions.

Element loses electrons → its oxidation number increases→ element is oxidized → oxidation reaction

Substance that contains the oxidized element is call thereducing agent.

Substance that contains the reduced element is call theoxidizing agent.

Element gains electrons → its oxidation number decreases→ element is reduced → reduction reaction

PbO (s) + CO (g) → Pb (s) + CO2 (g)

Cu − 2e− Cu2+

Problem Set 8

concentrations: mass percent and molarity;precipitation reactions; redox reactions.