1. Chemistry: Atoms First Julia Burdge & Jason Overby
Chapter 8.3 and 8.4
Book Homework: 25, 27, 29, 31, 33, 37, 43, 47, 49, 51, 53, 55 and 57
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. How to “Read” Chemical Equations
2 Mg + O MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7

3. 3 ways of representing the reaction of H2 with O2 to form H2O
A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products

4. Chemical Equations
8.1
A chemical equation uses chemical symbols to denote what occurs in a chemical reaction.
NH3 + HCl → NH4Cl
Ammonia and hydrogen chloride react to produce ammonium chloride.
Each chemical species that appears to the left of the arrow is called a reactant.
Each species that appears to the right of the arrow is called a product.

5. Interpreting and Writing Chemical Equations
Labels are used to indicate the physical state:
(g) gas
(l) liquid
(s) solid
(aq) aqueous [dissolved in water]
NH3(g) + HCl(g) → NH4Cl(s)
SO3(g) + H2O(l) → H2SO4(aq)

6. Balancing Chemical Equations
Chemical equations must be balanced so that the law of conservation of mass is obeyed.
Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas.

7. Balancing Chemical Equations
Generally, it will facilitate the balancing process if you do the following:
Change the coefficients of compounds before changing the coefficients of elements.
Treat polyatomic ions that appear on both sides of the equation as units.
Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient.

8. Calculations with Balanced Chemical Equations
8.3
Balanced chemical equations are used to predict how much product will form from a given amount of reactant.
2 moles of CO combine with 1 mole of O2 to produce 2 moles of CO2.
2 moles of CO is stoichiometrically equivalent to 2 moles of CO2.

9. Calculations with Balanced Chemical Equations
Consider the complete reaction of 3.82 moles of CO to form CO2. Calculate the number of moles of CO2 produced.

10. Calculations with Balanced Chemical Equations
Consider the complete reaction of 3.82 moles of CO to form CO2. Calculate the number of moles of O2 needed.

11. Mass Changes in Chemical Reactions
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units
3.8

12. Stoichiometric Ratio Coefficient B Moles B = X Moles A Coefficient A
The stoichiometric ratio is used to determine amounts of compounds
consumed or produced in a balanced chemical reaction

13. Worked Example 8.5
Urea [(NH2)2CO] is a by-product of protein metabolism. This waste product is formed in the liver and then filtered from the blood and excreted in the urine by the kidneys. Urea can be synthesized in the laboratory by the combination of ammonia and carbon dioxide according to the equation
(a) Calculate the amount of urea that will be produced by the complete reaction of 5.25 moles of ammonia. (b) Determine the stoichiometric amount of carbon dioxide required to react with 5.25 moles of ammonia.
Strategy Use the balanced chemical equation to determine the correct stoichiometric conversion factors, and then multiply by the number of moles of ammonia given.

14. Worked Example 8.5 (cont.) Solution
(a) moles (NH2)2CO produced = 5.25 mol NH3 ×
(b) moles CO2 produced = 5.25 mol NH3 ×
1 mol (NH2)2CO
2 mol NH3
= 2.63 mol
(NH2)2CO
1 mol CO2
2 mol NH3
= 2.63 mol CO2
Think About It As always, check to be sure that units cancel properly in the calculation. Also, the balanced equation indicates that there will be fewer moles of urea produced than ammonia consumed. Therefore, your calculated number of moles of urea (2.63) should be smaller than the number of moles given in the problem (5.25). Similarly, the stoichiometric coefficients in the balanced equation are the same for carbon dioxide and urea, so your answers to this problem should also be the same for both species.

15. Worked Example 8.6
Dinitrogen monoxide (N2O), also known as nitrous oxide or “laughing gas,” is used as an anesthetic in dentistry. It is manufactured by heating ammonium nitrate. The balanced equation is
NH4NO3(s) → N2O(g) + 2H2O(l)
(a) Calculate the mass of ammonium nitrate that must be heated in order to produce 10.0 g of nitrous oxide. (b) Determine the corresponding mass of water produced in the reaction. Δ
Strategy For part (a), use the molar mass of nitrous oxide to convert the given mass of nitrous oxide to moles, use the appropriate stoichiometric conversion factor to convert to moles of ammonium nitrate, and then use the molar mass of ammonium nitrate to convert to grams of ammonium nitrate. For part (b), use the molar mass of nitrous oxide to convert the given mass of nitrous oxide to moles, use the stoichiometric conversion factor to convert from moles of nitrous oxide to moles of water, and then use the molar mass of water to convert to grams of water.

16. Worked Example 8.6 (cont.) Solution (a) 10.0 g N2O × 1 mol N2O
0.227 mol NH4NO3 ×
Thus, 18.2 g of ammonium nitrate must be heated in order to produce 10.0 g of nitrous oxide.
(b) Starting with the number of moles of nitrous oxide determined in the first step of (a),
0.454 mol H2O ×
1 mol N2O
44.02 g N2O
= mol N2O
1 mol NH4NO3
1 mol N2O
Think About It Use the law of conservation of mass to check your answers. Make sure that the combined mass of both products is equal to the mass of reactant you determined in part (a). In this case (rounded to the appropriate number of significant figures), 10.0 g g = 18.2 g. Remember that small differences may arise as the result of rounding.
= mol NH4NO3
80.05 g NH4NO3
1 mol NH4NO3
= 18.2 g NH4NO3
2 mol H2O
1 mol N2O
= mol H2O
18.02 g H2O
1 mol H2O
= 8.18 g H2O

17. CO(g) + 2H2(g) → CH3OH(l) 8.4 Limiting Reactants
The reactant used up first in a reaction is called the limiting reactant.
Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant.
CO(g) + 2H2(g) → CH3OH(l)

18. Limiting Reactants
Consider the reaction between 5 moles of CO and 8 moles of H2 to produce methanol.
How many moles of H2 are necessary in order for all the CO to react?
How many moles of CO are necessary in order for all of the H2 to react?
10 moles of H2 required; 8 moles of H2 available; limiting reactant.
4 moles of CO required; 5 moles of CO available; excess reactant.
CO(g) + 2H2(g) → CH3OH(l)

19. 3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
Worked Example 8.7
Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7) react to form carbon dioxide gas, among other products.
3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
The formation of CO2 causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains g of sodium bicarbonate and g citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO2 forms.
Strategy Convert each of the reactant masses to moles. Use the balanced equation to write the stoichiometric conversion factor and determine which reactant is limiting. Next, determine the number of moles of excess reactant remaining and the number of moles of CO2 produced. Finally, use the appropriate molar masses to convert moles of excess reactant and moles of CO2 to grams.

20. Worked Example 8.7 (cont.) Solution 1.700 g NaHCO3 × 1 mol NaHCO3
1.000 g H3C6H5O7 ×
(a) To determine which reactant is limiting, calculate the amount of citric acid necessary to react completely with mol sodium bicarbonate.
mol NaHCO3×
The amount of H3C6H5O7 required to react with mol of NaHCO3 is more than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant.
1 mol NaHCO3
84.01 g NaHCO3
= mol NaHCO3
1 mol H3C6H5O7
g H3C6H5O7
= mol H3C6H5O7
1 mol H3C6H5O7
3 mol NaHCO3
= mol H3C6H5O7

21. Worked Example 8.7 (cont.) Solution
(b) To determine the mass of excess reactant (NaHCO3) left over, first calculate the amount of NaHCO3 that will react:
mol H3C6H5O7 ×
Thus, mole of NaHCO3 will be consumed, leaving mole unreacted. Convert the unreacted amount to grams as follows:
mol NaHCO3 ×
3 mol NaHCO3
1 mol H3C6H5O7
= mol NaHCO3
84.01 g NaHCO3
1 mol NaHCO3
= g NaHCO3

22. Worked Example 8.7 (cont.) Solution
(c) To determine the mass of CO2 produced, first calculate the moles of CO2 produced from the number of moles of limiting reactant (H3C6H5O7) consumed.
mol H3C6H5O7 ×
Convert this amount to grams as follows:
mol CO2×
To summarize the results: (a) citric acid is the limiting reactant, (b) g sodium bicarbonate remains unreacted, and (c) g carbon dioxide is produced.
Think About It In a problem such as this, it is a good idea to check your work by calculating the amounts of the other products in the reaction. According to the law of conservation of mass, the combined starting mass of the two reactants (1.700 g g = g) should equal the sum of the masses of products and leftover excess reactant. In this case, the masses of H2O and Na3C6H5O7 produced are g and g, respectively. The mass of CO2 produced is g [from part (c)] and the amount of excess NaHCO3 is g [from part (b)]. The total, g g g g, is g, identical to the total mass of reactants.
3 mol CO2
1 mol H3C6H5O7
= mol CO2
44.01 g CO2
1 mol CO2
= g CO2

23. Limiting Reagent
The reactant that is completely consumed by the reaction
The number of bicycles that can be assembled is limited by whichever part
runs out first. In the inventory shown in this figure, wheels are that part.

24. Limiting Reactants
The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product.
The actual yield is the amount of product actually obtained from a reaction.
The percent yield tells what percentage the actual yield is of the theoretical yield.

25. Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100
3.10

26. Worked Example 8.8
Aspirin, acetylsalicylic acid (C9H8O4), is the most commonly used pain reliever in the world. It is produced by the reaction of salicylic acid (C7H6O3) and acetic anhydride (C4H6O3) according to the following equation:
In a certain aspirin synthesis, g of salicylic acid and g of acetic anhydride are combined. Calculate the percent yield if g of aspirin are produced.
C7H6O3
salicylic acid +
C4H6O3
acetic anhydride →
C9H8O4
acetylsalicylic acid +
HC2H3O2
acetic acid
Strategy Convert reactant grams to moles, and determine which is the limiting reactant. Use the balanced equation to determine the moles of aspirin that can be produced and convert to grams for the theoretical yield. Use this and the actual yield given to calculate the percent yield.

27. Worked Example 8.8 (cont.) Solution 104.8 g C7H6O3× 1 mol C7H6O3
Because the two reactants combine in a 1:1 mole ratio, the reactant present in the smallest number of moles (in this case, salicylic acid) is the limiting reactant. According to the balanced equation, one mole of aspirin is produced for every mole of salicylic acid consumed.
Therefore, the theoretical yield of aspirin is mol. We convert this to grams using the molar mass of aspirin:
mol C9H8O4×
1 mol C7H6O3
g C7H6O3
= mol C7H6O3
1 mol C4H6O3
g C4H6O3
= mol C4H6O3
g C9H8O4
1 mol C9H8O4
= g C9H8O4

28. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O
3.8