ch 4. chemical quantities and aqueous reactions. ch 4 (g) + 2o 2 (g) co 2 (g) + 2h 2 o (g) 1 mol2...
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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
Can calculate how much other chemicals are requiredor produced if the amount of one chemical is known.
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)
5.02 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
5.02 mol x y z u
3
2
x
mol 5.02
mol 3
mol 2 mol 7.53x
y
mol 5.02
1
2 mol 2.51 y
z
mol 5.02
3
2 mol 7.53z
u
mol 5.02
3
2 mol 7.53u
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 131 ― 132
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 generated 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 and CH4 will have leftover: excess reagent.
= 0.5 molconsumed: 1 molx
mol 1
x
2
1 x = 0.5 mol
final: y z
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
is 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
2NH3(g) + 3CuO(s) N2(g) + 3Cu(s) + 3H2O(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
68.5 g CO reacts with 8.60 g H2 in the following reaction.
What is the limiting reagent? How many grams of excess
reagent is leftover? What is the theoretical yield of CH3OH?
If 35.7 g CH3OH is actually produced, what is the percent
yield of CH3OH?
H2(g) + CO(g) CH3OH(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
1.50 g of ammonia reacts with 2.75 g of oxygen gas to produce nitrogen monoxide and water.
NH3(g) + O2(g) NO(g) + H2O(g)
a) Balance the equation.b) What is the mass of O2 in grams required by NH3?c) Which reactant is the limiting reagent? d) How many grams of NO will be produced in theory?e) How many grams of H2O will be produced in theory?f) How many grams of the excess reagent remain unreacted?g) If only 1.80 g of NO are produced, what is the percent yield?
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
C12H22O11
electrolytes
nonelectrolytes
Based on the electrical conductivity in aqueous solution
strong electrolytes
weak electrolytes
salts
strong acids
strong bases
weak acids
weak basessolutes
strong electrolytes: dissociate 100 % into ions
weak electrolytes: only a small fraction dissociate into ions
nonelectrolytes: no dissociation
salts: NaCl, Na2SO4, Fe(ClO4)3 ……
strong acids: HCl, HNO3, H2SO4, HClO4
strong bases: NaOH, KOH
Base: compounds that give OH− when dissolved in water.
weak acids: acetic acid: HC2H3O2
weak bases: ammonia: NH3
remember
remember
HCl is Completely Ionized
Aqueous Solution of NaOH
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
0.50 mol of KBr is dissolved in water and forms a solution of 12 L. What is the molarity of the solution?
Example 4.5, page 141
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 142
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 143
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 144
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)
Fe(NO3)3(aq) + KOH(aq)
Fe(NO3)3(aq) Fe3+(aq) + 3NO3−(aq)
KOH(aq) K+ (aq) + OH− (aq)
Fe3+(aq) + 3NO3−(aq) +3K+ (aq) +3OH− (aq) Fe(OH)3(s) + 3NO3
−(aq) + 3K+(aq)
Fe3+(aq) + 3OH− (aq) Fe(OH)3(s)
3KOH(aq) 3K+ (aq) + 3OH− (aq)
Fe(NO3)3(aq) + 3KOH(aq) Fe(OH)3(s) + 3KNO3(aq)
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
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 158 and the online instructionfor next week’s titration
Example 4.14
The titration of a 10.00-mL sample of an HCl solution of unknown concentration requires 12.54 mL of a 0.100 M NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution in M?
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 numbers (states)
1) For atoms in its elemental form, oxidation number = 0
A way to keep track of the electrons gained or lost
Na, Ag, Ar, O2, N2, P4
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
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−