Ch 4. Chemical Quantities and Aqueous Reactions
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Transcript of Ch 4. Chemical Quantities and Aqueous Reactions
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Ch 4. Chemical Quantities and Aqueous Reactions
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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.
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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
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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?
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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
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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
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Other Examples: page 143 ― 144
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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.
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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.
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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.
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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=
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++
+++ +
+ +
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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
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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)
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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
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Problem Set 7
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Review problem sets 6 & 7
Review experiment 11
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Midterm Exam
Time: next week during lab session
Material covered: up to this point
Problem sets are very helpful
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Classification of Matter
Matter
Elements
Compounds
Mixtures(multiple components)
Pure Substances(one component)
Homogeneous(visibly indistinguishable)
Heterogeneous (visibly distinguishable)
(Solutions)
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Solute + Solvent = Solution
Solvent = water, aqueous solution
Water can dissolve many substances
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O
H H
H2O
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Solution conducts electricity well
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C12H22O11
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Solution does not conduct electricity
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Solution conducts electricity, but weakly
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electrolytes
nonelectrolytes
Based on the electrical conductivity in aqueous solution
strong electrolytes
weak electrolytessolutes
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strong electrolytes: dissociate 100 % into ions
weak electrolytes: only a small fraction dissociate into ions
nonelectrolytes: no dissociation
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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
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Reaction of NH3 in Water
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concentrations
% 100 sample wholeof mass
component of mass percent mass
% 100 solution of mass
solute of mass percent mass
no unit
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10. g of sugar is dissolved in 40. g of water.
What is the mass percent of sugar in this solution?
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moles of soluteMolarity (M)
liters of solution
Unit: mol/L or M
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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
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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)
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Solution Dilution
Concentrated solutions for storage, called stock solutions
stock solution + water desired solution
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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?
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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?
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Types of reactions
Precipitation reactions
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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
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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
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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)
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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)
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Types of reactions
Precipitation reactions
Acid-base reactions
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Acid: Substance that produces H+ ions in aqueous solution
Base: Substance that produces OH− ions in aqueous solution
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Try to remember them
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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
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HCl(aq) H+(aq) + Cl−(aq)
What is the molarity ofHCl(aq) or H+(aq)?
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When reaction completes
nNaOH = nHCl
MNaOHVNaOH = MHClVHCl
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
prepared,known
measuredby buret, known
unknown measured bypippet, known
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MNaOHVNaOH = MHClVHCl
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Read acid-base titration starting on page 171.Read the online instruction next week for the titration experiment.
End point: light pink
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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)
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Types of reactions
Precipitation reactions
Acid-base reactions
Oxidation-Reduction reactions
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Reactions that involve electron transfer are called
oxidation-reduction reactions, or redox reactions.
2Mg(s) + O2(g) → 2MgO(s)
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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
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O
H H
H2O
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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−
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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
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PbO (s) + CO (g) → Pb (s) + CO2 (g)
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Cu − 2e− Cu2+
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Problem Set 8
concentrations: mass percent and molarity;precipitation reactions; redox reactions.