1. Chapter 4 Chemical Quantities and Aqueous Reactions
Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 4 Chemical Quantities and Aqueous Reactions
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall

2. Reaction Stoichiometry
the numerical relationships between chemical amounts in a reaction is called stoichiometry
the coefficients in a balanced chemical equation 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
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3. Predicting Amounts from Stoichiometry
the amounts of any other substance in a chemical reaction can be determined from the amount of just one substance
How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18?
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 moles C8H18 : 16 moles CO2
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4. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline
assuming that gasoline is octane, C8H18, the equation for the reaction is:
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
the equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the Concept Plan will be:
g C8H18
mol CO2
g CO2
mol C8H18
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5. 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2
Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline
Given:
Find:
3.4 x 1015 g C8H18
g CO2
Concept Plan:
Relationships:
1 mol C8H18 = g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2
g C8H18
mol CO2
g CO2
mol C8H18
Solution:
Check:
since 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

6. C6H12O6(s) + 6 O2(g) ® 6 CO2(g) + 6 H2O(l)
Practice
According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose?
C6H12O6(s) + 6 O2(g) ® 6 CO2(g) + 6 H2O(l)
convert 9.0 g of glucose into moles (MM 180)
convert moles of glucose into moles of water
convert moles of water into grams (MM 18.02)
convert grams of water into mL
How? what is the relationship between mass and volume?
density of water = 1.00 g/mL
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7. C6H12O6(s) + 6 O2(g) ® 6 CO2(g) + 6 H2O(l)
Practice
According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose?
C6H12O6(s) + 6 O2(g) ® 6 CO2(g) + 6 H2O(l)
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8. Limiting Reactant
for reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others
when this reactant is used up, the reaction stops and no more product is made
the reactant that limits the amount of product is called the limiting reactant
sometimes called the limiting reagent
the limiting reactant gets completely consumed
reactants not completely consumed are called excess reactants
the amount of product that can be made from the limiting reactant is called the theoretical yield
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9. Things Don’t Always Go as Planned!
many things can happen during the course of an experiment that cause the loss of product
the amount of product that is made in a reaction is called the actual yield
generally less than the theoretical yield, never more!
the efficiency of product recovery is generally given as the percent yield
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10. Limiting and Excess Reactants in 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 1 molecule of CH4 reacts with 2 molecules of O2 H C + O
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11. Limiting and Excess Reactants in the Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
If we have 5 molecules of CH4 and 8 molecules of O2, which is the limiting reactant? H C + O ?

12. Limiting and Excess Reactants in the Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) H C + O
since less CO2 can be made from the O2 than the CH4, the O2 is the limiting reactant

13. Example 4.4 Finding Limiting Reactant, Theoretical Yield, and Percent Yield

14. Example:
When 28.6 kg of C are allowed to react with 88.2 kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield.
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15. Given: 28.6 kg C 88.2 kg TiO2 42.8 kg Ti produced
Example: When 28.6 kg of C reacts with kg of TiO2, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Write down the given quantity and its units.
Given: 28.6 kg C
88.2 kg TiO2
42.8 kg Ti produced
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16. Find: limiting reactant theoretical yield percent yield
Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: kg C, 88.2 kg TiO2, 42.8 kg Ti
Write down the quantity to find and/or its units.
Find: limiting reactant
theoretical yield
percent yield
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17. Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
Write a Concept Plan: }
smallest
amount is
from
limiting
reactant kg
TiO2 C g
TiO2 C
mol C
TiO2
mol Ti
smallest
mol Ti
g Ti
kg Ti
T.Y.
% Yield

18. Collect Needed Relationships:
Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Collect Needed Relationships:
1000 g = 1 kg
Molar Mass TiO2 = g/mol
Molar Mass Ti = g/mol
Molar Mass C = g/mol
1 mole TiO2 : 1 mol Ti (from the chem. equation)
2 mole C  1 mol Ti (from the chem. equation)

19. Apply the Concept Plan:
Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the Concept Plan:
Limiting Reactant
smallest moles of Ti
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20. Apply the Concept Plan:
Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the Concept Plan:
Theoretical Yield
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21. Apply the Concept Plan:
Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the Concept Plan:
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22. Limiting Reactant = TiO2 Theoretical Yield = 52.9 kg
Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Check the Solutions:
Limiting Reactant = TiO2
Theoretical Yield = 52.9 kg
Percent Yield = 80.9%
Since Ti has lower molar mass than TiO2, the T.Y. makes sense
The Percent Yield makes sense as it is less than 100%.
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23. Practice – How many grams of N2(g) can be made from 9
Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
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24. Practice – How many grams of N2(g) can be made from 9
Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
Given:
Find:
9.05 g NH3, 45.2 g CuO
g N2
Concept Plan:
Relationships:
1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = g
2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2
g NH3
mol N2
mol NH3
g CuO
mol N2
mol CuO
g N2
smallest moles N2

25. Practice – How many grams of N2(g) can be made from 9
Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
Solution:
Check:
units are correct, and since there are fewer moles of N2 than CuO in the reaction and N2 has a smaller mass, the number makes sense

26. Solutions
when table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous
the salt is still there, as you can tell from the taste, or simply boiling away the water
homogeneous mixtures are called solutions
the component of the solution that changes state is called the solute
the component that keeps its state is called the solvent
if both components start in the same state, the major component is the solvent
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27. Describing Solutions
since solutions are mixtures, the composition can vary from one sample to another
pure substances have constant composition
salt water samples from different seas or lakes have different amounts of salt
so to describe solutions accurately, we must describe how much of each component is present
we saw that with pure substances, we can describe them with a single name because all samples identical
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28. Solution Concentration
qualitatively, solutions are often described as dilute or concentrated
dilute solutions have a small amount of solute compared to solvent
concentrated solutions have a large amount of solute compared to solvent
quantitatively, the relative amount of solute in the solution is called the concentration
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29. Solution Concentration Molarity
moles of solute per 1 liter of solution
used because it describes how many molecules of solute in each liter of solution
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30. Preparing 1 L of a 1.00 M NaCl Solution
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31. since most solutions are between 0 and 18 M, the answer makes sense
Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution
Sort Information
Given:
Find:
25.5 g KBr, 1.75 L solution
Molarity, M
Strategize
Concept Plan:
Relationships:
1 mol KBr = g,
M = moles/L
g KBr
mol KBr
L sol’n M
Follow the Concept Plan to Solve the problem
Solution:
Check
Check:
since most solutions are between 0 and 18 M, the answer makes sense

32. Using Molarity in Calculations
molarity shows the relationship between the moles of solute and liters of solution
If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar
2 liters = 4.0 moles sugar
0.5 liters = 1.0 mole sugar
1 L solution : 2 moles sugar
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33. Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH?
Sort Information
Given:
Find:
0.125 M NaOH, mol NaOH
liters, L
Strategize
Concept Plan:
Relationships:
0.125 mol NaOH = 1 L solution
mol NaOH
L sol’n
Follow the Concept Plan to Solve the problem
Solution:
Check
Check:
since each L has only mol NaOH, it makes sense that mol should require a little more than 2 L

34. moles solute in solution 1 = moles solute in solution 2
Dilution
often, solutions are stored as concentrated stock solutions
to make solutions of lower concentrations from these stock solutions, more solvent is added
the amount of solute doesn’t change, just the volume of solution
moles solute in solution 1 = moles solute in solution 2
the concentrations and volumes of the stock and new solutions are inversely proportional
M1∙V1 = M2∙V2
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35. Example 4. 7 – To what volume should you dilute 0. 200 L of 15
Example 4.7 – To what volume should you dilute L of 15.0 M NaOH to make 3.00 M NaOH?
Sort Information
Given:
Find:
V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Strategize
Concept Plan:
Relationships:
M1V1 = M2V2
V1, M1, M2 V2
Follow the Concept Plan to Solve the problem
Solution:
Check
Check:
since the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does

36. Solution Stoichiometry
since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction
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37. Example 4. 8 – What volume of 0
Example 4.8 – What volume of M KCl is required to completely react with L of M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq)
Sort Information
Given:
Find:
0.150 M KCl, L of M Pb(NO3)2
L KCl
Strategize
Concept Plan:
Relationships:
1 L Pb(NO3)2 = mol, 1 L KCl = mol,
1 mol Pb(NO3)2 = 2 mol KCl
L Pb(NO3)2
mol KCl
L KCl
mol Pb(NO3)2
Follow the Concept Plan to Solve the problem
Solution:
Check
Check:
since need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x volume Pb(NO3)2

38. What Happens When a Solute Dissolves?
there are attractive forces between the solute particles holding them together
there are also attractive forces between the solvent molecules
when we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules
if the attractions between solute and solvent are strong enough, the solute will dissolve

39. Table Salt Dissolving in Water
Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal
When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions
The result is a solution with free moving charged particles able to conduct electricity
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40. Electrolytes and Nonelectrolytes
materials that dissolve in water to form a solution that will conduct electricity are called electrolytes
materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes
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41. Molecular View of Electrolytes and Nonelectrolytes
in order to conduct electricity, a material must have charged particles that are able to flow
electrolyte solutions all contain ions dissolved in the water
ionic compounds are electrolytes because they all dissociate into their ions when they dissolve
nonelectrolyte solutions contain whole molecules dissolved in the water
generally, molecular compounds do not ionize when they dissolve in water
the notable exception being molecular acids
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42. Salt vs. Sugar Dissolved in Water
molecular compounds do not dissociate when they dissolve
ionic compounds dissociate into ions when they dissolve
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43. Acids
acids are molecular compounds that ionize when they dissolve in water
the molecules are pulled apart by their attraction for the water
when acids ionize, they form H+ cations and anions
the percentage of molecules that ionize varies from one acid to another
acids that ionize virtually 100% are called strong acids
HCl(aq)  H+(aq) + Cl-(aq)
acids that only ionize a small percentage are called weak acids
HF(aq)  H+(aq) + F-(aq)
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44. Strong and Weak Electrolytes
strong electrolytes are materials that dissolve completely as ions
ionic compounds and strong acids
their solutions conduct electricity well
weak electrolytes are materials that dissolve mostly as molecules, but partially as ions
weak acids
their solutions conduct electricity, but not well
when compounds containing a polyatomic ion dissolve, the polyatomic ion stays together
Na2SO4(aq)  2 Na+(aq) + SO42-(aq)
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
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45. Classes of Dissolved Materials
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46. Solubility of Ionic Compounds
some ionic compounds, like NaCl, dissolve very well in water at room temperature
other ionic compounds, like AgCl, dissolve hardly at all in water at room temperature
compounds that dissolve in a solvent are said to be soluble, while those that do not are said to be insoluble
NaCl is soluble in water, AgCl is insoluble in water
the degree of solubility depends on the temperature
even insoluble compounds dissolve, just not enough to be meaningful
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47. When Will a Salt Dissolve?
Predicting whether a compound will dissolve in water is not easy
The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results
we call this method the empirical method
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48. Solubility Rules Compounds that Are Generally Soluble in Water
Compounds Containing the Following Ions are Generally Soluble
Exceptions
(when combined with ions on the left the compound is insoluble)
Li+, Na+, K+, NH4+
none
NO3–, C2H3O2–
Cl–, Br–, I–
Ag+, Hg22+, Pb2+
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
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49. Solubility Rules Compounds that Are Generally Insoluble
Compounds Containing the Following Ions are Generally Insoluble
Exceptions
(when combined with ions on the left the compound is soluble or slightly soluble)
OH–
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
S2–
CO32–, PO43–
Li+, Na+, K+, NH4+
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50. Precipitation Reactions
reactions between aqueous solutions of ionic compounds that produce an ionic compound that is insoluble in water are called precipitation reactions and the insoluble product is called a precipitate
picture from Tro Intro Chem, 2nd Ed.
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51. 2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2 KNO3(aq)

52. No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
all ions still present,  no reaction
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53. Process for Predicting the Products of a Precipitation Reaction
Determine what ions each aqueous reactant has
Determine formulas of possible products
Exchange ions
(+) ion from one reactant with (-) ion from other
Balance charges of combined ions to get formula of each product
Determine Solubility of Each Product in Water
Use the solubility rules
If product is insoluble or slightly soluble, it will precipitate
If neither product will precipitate, write no reaction after the arrow
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54. Process for Predicting the Products of a Precipitation Reaction
If either product is insoluble, write the formulas for the products after the arrow – writing (s) after the product that is insoluble and will precipitate, and (aq) after products that are soluble and will not precipitate
Balance the equation
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55. Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride
Write the formulas of the reactants
K2CO3(aq) + NiCl2(aq) 
Determine the possible products
Determine the ions present
(K+ + CO32-) + (Ni2+ + Cl-) 
Exchange the Ions
(K+ + CO32-) + (Ni2+ + Cl-)  (K+ + Cl-) + (Ni2+ + CO32-)
Write the formulas of the products
cross charges and reduce
K2CO3(aq) + NiCl2(aq)  KCl + NiCO3

56. does not apply since NiCO3 is insoluble
Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride
Determine the solubility of each product
KCl is soluble
NiCO3 is insoluble
If both products soluble, write no reaction
does not apply since NiCO3 is insoluble
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57. Write (aq) next to soluble products and (s) next to insoluble products
Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride
Write (aq) next to soluble products and (s) next to insoluble products
K2CO3(aq) + NiCl2(aq)  KCl(aq) + NiCO3(s)
Balance the Equation
K2CO3(aq) + NiCl2(aq)  2 KCl(aq) + NiCO3(s)
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58. 2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s)
Ionic Equations
equations which describe the chemicals put into the water and the product molecules are called molecular equations
2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s)
equations which describe the actual dissolved species are called complete ionic equations
aqueous strong electrolytes are written as ions
soluble salts, strong acids, strong bases
insoluble substances, weak electrolytes, and nonelectrolytes written in molecule form
solids, liquids, and gases are not dissolved, therefore molecule form
2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq) ® 2K+1(aq) + 2NO3-1(aq) + Mg(OH)2(s)
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59. Ionic Equations
ions that are both reactants and products are called spectator ions
2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq) ® 2K+1(aq) + 2NO3-1(aq) + Mg(OH)2(s)
an ionic equation in which the spectator ions are
removed is called a net ionic equation
2OH-1(aq) + Mg+2(aq) ® Mg(OH)2(s)
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60. 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)
as long as the salt that forms is soluble in water
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61. Acids and Bases in Solution
acids ionize in water to form H+ ions
more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+
most chemists use H+ and H3O+ interchangeably
bases dissociate in water to form OH ions
bases, like NH3, that do not contain OH ions, produce OH by pulling H off water molecules
in the reaction of an acid with a base, the H+ from the acid combines with the OH from the base to make water
the cation from the base combines with the anion from the acid to make the salt
acid + base salt + water
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62. Common Acids

63. Common Bases
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64. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
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65. Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq) 
Determine the possible products
Determine the ions present when each reactant dissociates
(H+ + NO3-) + (Ca+2 + OH-) 
Exchange the ions, H+1 combines with OH-1 to make H2O(l)
(H+ + NO3-) + (Ca+2 + OH-)  (Ca+2 + NO3-) + H2O(l)
Write the formula of the salt
cross the charges
(H+ + NO3-) + (Ca+2 + OH-)  Ca(NO3)2 + H2O(l)
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66. Determine the solubility of the salt
Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
Determine the solubility of the salt
Ca(NO3)2 is soluble
Write an (s) after the insoluble products and a (aq) after the soluble products
HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + H2O(l)
Balance the equation
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
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67. Eliminate spectator ions to get net-ionic equation
Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
Dissociate all aqueous strong electrolytes to get complete ionic equation
not H2O
2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq)  Ca+2(aq) + 2 NO3-(aq) + H2O(l)
Eliminate spectator ions to get net-ionic equation
2 H+1(aq) + 2 OH-1(aq)  2 H2O(l)
H+1(aq) + OH-1(aq)  H2O(l)
Tro, Chemistry: A Molecular Approach

68. Titration
often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration
in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio
the unknown solution is added slowly from an instrument called a burette
a long glass tube with precise volume markings that allows small additions of solution
Tro, Chemistry: A Molecular Approach

69. Acid-Base Titrations
the difficulty is determining when there has been just enough titrant added to complete the reaction
the titrant is the solution in the burette
in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity
the chemical is called an indicator
at the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH
aka the equivalence point
Tro, Chemistry: A Molecular Approach

70. Titration
Tro, Chemistry: A Molecular Approach

71. Titration The base solution is the titrant in the burette.
As the base is added to
the acid, the H+ reacts with
the OH– to form water.
But there is still excess
acid present so the color
does not change.
At the titration’s endpoint,
just enough base has been
added to neutralize all the
acid. At this point the
indicator changes color.
Tro, Chemistry: A Molecular Approach

72. 12.54 mL of 0.100 M NaOH Given: 10.00 mL HCl
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Write down the given quantity and its units.
Given: mL HCl
12.54 mL of M NaOH
Tro, Chemistry: A Molecular Approach

73. Find: concentration HCl, M
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Write down the quantity to find, and/or its units.
Find: concentration HCl, M
Tro, Chemistry: A Molecular Approach

74. Collect Needed Equations and Conversion Factors:
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Collect Needed Equations and Conversion Factors:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
 1 mole HCl = 1 mole NaOH
0.100 M NaOH 0.100 mol NaOH  1 L sol’n
Tro, Chemistry: A Molecular Approach

75. Example 4. 14: The titration of 10
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Write a Concept Plan: mL
NaOH L
NaOH
mol
NaOH
mol
HCl mL
HCl L
HCl
Tro, Chemistry: A Molecular Approach

76. Apply the Solution Map:
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the Solution Map:
= 1.25 x 10-3 mol HCl
Tro, Chemistry: A Molecular Approach

77. Apply the Concept Plan:
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the Concept Plan:
Tro, Chemistry: A Molecular Approach

78. Check the Solution: HCl solution = 0.125 M
Example: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
CF: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Check the Solution:
HCl solution = M
The units of the answer, M, are correct.
The magnitude of the answer makes sense since
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.
Tro, Chemistry: A Molecular Approach

79. Gas Evolving Reactions
Some reactions form a gas directly from the ion exchange
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water
K2SO3(aq) + H2SO4(aq)  K2SO4(aq) + H2SO3(aq)
H2SO3  H2O(l) + SO2(g)
Tro, Chemistry: A Molecular Approach

80. NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
Tro, Chemistry: A Molecular Approach

81. Compounds that Undergo Gas Evolving Reactions
Reactant
Type
Reacting
With
Ion Exchange
Product
Decom-pose?
Gas
Formed
Example
metalnS,
metal HS
acid
H2S no
K2S(aq) + 2HCl(aq) 
2KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
H2CO3
yes
CO2
K2CO3(aq) + 2HCl(aq) 
2KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
H2SO3
SO2
K2SO3(aq) + 2HCl(aq) 
2KCl(aq) + SO2(g) + H2O(l)
(NH4)nanion
base
NH4OH
NH3
KOH(aq) + NH4Cl(aq) 
KCl(aq) + NH3(g) + H2O(l)
Tro, Chemistry: A Molecular Approach

82. Example When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves
Write the formulas of the reactants
Na2CO3(aq) + HNO3(aq) 
Determine the possible products
Determine the ions present when each reactant dissociates
(Na+1 + CO3-2) + (H+1 + NO3-1) 
Exchange the anions
(Na+1 + CO3-2) + (H+1 + NO3-1)  (Na+1 + NO3-1) + (H+1 + CO3-2)
Write the formula of compounds
cross the charges
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
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83. Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
Example When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves
Check to see either product H2S - No
Check to see if either product decomposes – Yes
H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
Tro, Chemistry: A Molecular Approach

84. Determine the solubility of other product
Example When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves
Determine the solubility of other product
NaNO3 is soluble
Write an (s) after the insoluble products and a (aq) after the soluble products
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
Balance the equation
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3 + CO2(g) + H2O(l)
Tro, Chemistry: A Molecular Approach

85. Other Patterns in Reactions
the precipitation, acid-base, and gas evolving reactions all involved exchanging the ions in the solution
other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions
also known as redox reactions
many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
Tro, Chemistry: A Molecular Approach

86. Combustion as Redox 2 H2(g) + O2(g)  2 H2O(g)
Tro, Chemistry: A Molecular Approach

87. Redox without Combustion 2 Na(s) + Cl2(g)  2 NaCl(s)
2 Na  2 Na+ + 2 e
Cl2 + 2 e  2 Cl
Tro, Chemistry: A Molecular Approach

88. Reactions of Metals with Nonmetals
consider the following reactions:
4 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
the reaction involves a metal reacting with a nonmetal
in addition, both reactions involve the conversion of free elements into ions
4 Na(s) + O2(g) → 2 Na+2O– (s)
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Tro, Chemistry: A Molecular Approach

89. Oxidation and Reduction
in order to convert a free element into an ion, the atoms must gain or lose electrons
of course, if one atom loses electrons, another must accept them
reactions where electrons are transferred from one atom to another are redox reactions
atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced
Leo
Ger
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na → Na+ + 1 e– oxidation
Cl2 + 2 e– → 2 Cl– reduction
Tro, Chemistry: A Molecular Approach

90. Electron Bookkeeping
for reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the electrons are transferred
chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction
even though they look like them, oxidation states are not ion charges!
oxidation states are imaginary charges assigned based on a set of rules
ion charges are real, measurable charges
Tro, Chemistry: A Molecular Approach

91. Rules for Assigning Oxidation States
rules are in order of priority
free elements have an oxidation state = 0
Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
monatomic ions have an oxidation state equal to their charge
Na = +1 and Cl = -1 in NaCl
(a) the sum of the oxidation states of all the atoms in a compound is 0
Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
Tro, Chemistry: A Molecular Approach

92. Rules for Assigning Oxidation States
(b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion
N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1
(a) Group I metals have an oxidation state of +1 in all their compounds
Na = +1 in NaCl
(b) Group II metals have an oxidation state of +2 in all their compounds
Mg = +2 in MgCl2
Tro, Chemistry: A Molecular Approach

93. Rules for Assigning Oxidation States
in their compounds, nonmetals have oxidation states according to the table below
nonmetals higher on the table take priority
Nonmetal
Oxidation State
Example F -1
CF4 H +1
CH4 O -2
CO2
Group 7A
CCl4
Group 6A
CS2
Group 5A -3
NH3
Tro, Chemistry: A Molecular Approach

94. Practice – Assign an Oxidation State to Each Element in the following
Br2 K+
LiF
CO2
SO42-
Na2O2
Tro, Chemistry: A Molecular Approach

95. Practice – Assign an Oxidation State to Each Element in the following
Br2 Br = 0, (Rule 1)
K+ K = +1, (Rule 2)
LiF Li = +1, (Rule 4a) & F = -1, (Rule 5)
CO2 O = -2, (Rule 5) & C = +4, (Rule 3a)
SO42- O = -2, (Rule 5) & S = +6, (Rule 3b)
Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a)
Tro, Chemistry: A Molecular Approach

96. Oxidation and Reduction Another Definition
oxidation occurs when an atom’s oxidation state increases during a reaction
reduction occurs when an atom’s oxidation state decreases during a reaction
CH O2 → CO2 + 2 H2O –
oxidation
reduction
Tro, Chemistry: A Molecular Approach

97. Oxidation–Reduction oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them
the reactant that reduces an element in another reactant is called the reducing agent
the reducing agent contains the element that is oxidized
the reactant that oxidizes an element in another reactant is called the oxidizing agent
the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
Tro, Chemistry: A Molecular Approach

98. Identify the Oxidizing and Reducing Agents in Each of the Following
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O
Tro, Chemistry: A Molecular Approach

99. Identify the Oxidizing and Reducing Agents in Each of the Following
red ag
ox ag
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O
reduction
oxidation
ox ag
red ag
reduction
oxidation
Tro, Chemistry: A Molecular Approach

100. Combustion Reactions
Reactions in which O2(g) is a reactant are called combustion reactions
Combustion reactions release lots of energy
Combustion reactions are a subclass of oxidation-reduction reactions
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
Tro, Chemistry: A Molecular Approach

101. Combustion Products
to predict the products of a combustion reaction, combine each element in the other reactant with oxygen
Reactant
Combustion Product
contains C
CO2(g)
contains H
H2O(g)
contains S
SO2(g)
contains N
NO(g) or NO2(g)
contains metal
M2On(s)
Tro, Chemistry: A Molecular Approach

102. Practice – Complete the Reactions
combustion of C3H7OH(l)
combustion of CH3NH2(g)
Tro, Chemistry: A Molecular Approach

103. Practice – Complete the Reactions
C3H7OH(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
CH3NH2(g) + 3 O2(g)  CO2(g) + 2 H2O(g) + NO2(g)
Tro, Chemistry: A Molecular Approach