1. Chapter 4 Skills Balance Equations for simple chemical reactions
Perform stoichiometry calculations using balanced chemical equations
Mass relationships
Amounts tables
Understand the impact of a limiting reactant on a chemical reaction
Calculate the theoretical and percent yields of a chemical reaction
Stoichiometric uses
Analysis of mixtures
Determination of empirical or molecular formulas
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Kull Chem 105 Chapter 2

2. Balance Equations for simple chemical reactions
4. Balance the following equations:
(a) Cr(s) + Cl2(g)  CrCl3(s)
(b) SiO2(s) + C(s)  Si(s) + CO(g)
(c) Fe(s) + H2O(g)  Fe3O4(s) + H2(g)
4.4 (a) 2 Cr(s) + 3 Cl2(g)  2 CrCl3(s) (b) SiO2(s) + 2 C(s)  Si(s) + 2 CO(g) (c) 3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)
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3. Perform stoichiometry calculations using balanced chemical equations
14. The formation of water-insoluble silver chloride is useful in the analysis of chloride-containing substances. Consider the following unbalanced equation:
BaCl2(aq) + AgNO3(aq)  AgCl(s) + Ba(NO3)2(aq)
(a) Write the balanced equation.
(b) What mass AgNO3, in grams, is required for complete reaction with g of BaCl2?
What mass of AgCl is produced?
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4. GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
Mass
product
Mass
reactant
Stoichiometric
factor
Moles
reactant
Moles
product
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5. Perform stoichiometry calculations using balanced chemical equations
11. Iron metal reacts with oxygen to give iron(III) oxide, Fe2O3.
(a) Write a balanced equation for the reaction.
(b) If an ordinary iron nail (assumed to be pure iron) has a mass of 2.68 g, what mass of Fe2O3, in grams, is produced if the nail is converted completely to the oxide?
(c) What mass of O2, in grams, is required for the
reaction?
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6. Answers
4.4 (a) 2 Cr(s) + 3 Cl2(g)  2 CrCl3(s) (b) SiO2(s) + 2 C(s)  Si(s) + 2 CO(g) (c) 3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)
4.14 (a) BaCl2(aq) + 2 AgNO3(aq) 
2 AgCl(s) + Ba(NO3)2(aq) (b) g BaCl2 · · · = g AgNO g BaCl2 · · · = g AgCl
4.11 (a)4 Fe(s) + 3 O2(g)  2 Fe2O3(s) (b) 2.68 g Fe · · · = 3.83 g Fe2O3 (c) 3.83 g Fe2O3– 2.68 g Fe = 1.15 g O2
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7. Chemical Equations
Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change.
Because of the principle of the Law of the Conservation of Matter , an equation must be balanced.
It must have the same number of atoms of the same kind on both sides.
Demo of conservation of matter, See Screen 4.3.
2HgO(s) ---> 2 Hg(liq) + O2(g)
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8. Chemical Analysis
Active Figure 4.8
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9. Limiting reactant Vinegar is acetic acid: CH3COOH
Baking soda is sodium bicarbonate: NaHCO3
Mixing the two is simply and acid base reaction.
CH3COOH + NaHCO3 ---> CH3COONa + H2CO3
(60.0 g/mole) (84 g/mole)
That last product is carbonic acid which quickly decomposes into carbon dioxide and water:
H2CO3 ---> H2O + CO2
The CO2 is what you see foaming and bubbling in this reaction.
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10. Understand the impact of a limiting reactant on a chemical reaction
24. Aluminum chloride, AlCl3, is made by treating scrap aluminum with chlorine.
2 Al(s) + 3 Cl2(g)  2 AlCl3(s)
If you begin with 2.70 g of Al and 4.05 g of Cl2,
(a) Which reactant is limiting?
(b) What mass of AlCl3 can be produced?
(c) What mass of the excess reactant remains when the reaction is completed?
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11. Answers 4.24 (a) 2.70 g Al · = 0.100 mol Al
4.06 g Cl2 · = mol Cl2 The required mole ratio is 2 mol Al to 3 mol Cl2. More Al is available than required, so Cl2 is the limiting reactant. (b) mol Cl2 · · = 5.09 g AlCl3 (c) mol Cl2 · · = 1.03 g Al used g Al available – 1.03 g Al used = 1.67 g Al remains Equation 2 Al(s) Cl2(g)  2 AlCl3(s)
Initial amount (mol) Change (mol) – – Amount after rxn (mol)
g CO2 · · · = g NaHCO · 100% = 43.5% NaHCO3
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12. Understand the impact of a limiting reactant on a chemical reaction
20. ■ Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide with ammonium chloride.
CaO(s) + 2 NH4Cl(s)  2 NH3(g) + H2O(g) +CaCl2(s)
If 112 g of CaO and 224 g of NH4Cl are mixed, what mass of NH3 can be produced?
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13. Answers
4.20
112 g CaO · = 2.00 mol CaO
224 g NH4Cl · = 4.19 mol NH4Cl  CaO has fewer moles and is the limiting reactant 112 g CaO · · · = 68.0 g NH3
Equation
CaO(s) + 2 NH4Cl(s)  2 NH3(g) + H2O(g) + CaCl2(s)
Initial (mol)
Change (mol) – –
Amt after rxn(mol)
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14. Calculate the theoretical and percent yields of a chemical reaction
30. A reaction studied by Wächtershäuser and Huber (see “Black Smokers and the Origins of Life”) is
2 CH3SH + CO  CH3COSCH3 + H2S
If you begin with 10.0 g of CH3SH, and excess CO,
(a) What is the theoretical yield of CH3COSCH ?
(b) If 8.65 g of CH3COSCH3 is isolated, what is its percent yield?
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15. Answers 4.30 (a) 10.0 g CH3SH · · · = 9.37 g CH3COSCH3
(b) · 100% = 92.3% yield
g CO2 · · · = g NaHCO · 100% = 43.5% NaHCO3
g mixture – g after heating = g H2O lost g H2O · · · = 2.41 g CuCl2·2 H2O · 100% = 91.4% CuCl2·2 H2O
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16. Chemical Analysis
An impure sample of the mineral thenardite contains Na2SO4.
Mass of mineral sample = g
The Na2SO4 in the sample is converted to insoluble BaSO4.
The mass of BaSO4 is g
What is the mass percent of Na2SO4 in the mineral?
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17. Chemical Analysis Na2SO4(aq) + BaCl2(aq) --> 2 NaCl(aq) + BaSO4(s)
0.177 g BaSO4 (1 mol/233.4 g) = 7.58 x 10-4 mol BaSO4
7.58 x 10-4 mol BaSO4 (1 mol Na2SO4/1 mol BaSO4) = 7.58 x 10-4 mol Na2SO4
7.58 x 10-4 mol Na2SO4 (142.0 g/1 mol) = g Na2SO4
(0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4
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18. Chemical Analysis
An impure sample of the mineral thenardite contains Na2SO4.
Sample mass = g
The Na2SO4 is converted to insoluble BaSO4.
The mass of BaSO4 is g
What is the mass percent of Na2SO4 in the mineral?
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19. Determining the Formula of a Hydrocarbon by Combustion
Active Figure 4.9
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20. Using Stoichiometry to Determine a Formula
Burn g of a hydrocarbon, CxHy, and produce g of CO2 and g of H2O.
CxHy + some oxygen ---> g CO g H2O
What is the empirical formula of CxHy?
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21. Using Stoichiometry to Determine a Formula
CxHy + some oxygen ---> g CO g H2O
First, recognize that all C in CO2 and all H in H2O is from CxHy.
0.379 g CO2
+O2
Puddle of CxHy
0.115 g
1 CO2 molecule forms for each C atom in CxHy
g H2O
1 H2O molecule forms for each 2 H atoms in CxHy
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22. Using Stoichiometry to Determine a Formula
CxHy + some oxygen ---> g CO g H2O
First, recognize that all C in CO2 and all H in H2O is from CxHy.
1. Calculate amount of C in CO2
8.61 x 10-3 mol CO2 --> 8.61 x 10-3 mol C
2. Calculate amount of H in H2O
5.744 x 10-3 mol H2O -- >1.149 x 10-2 mol H
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23. Using Stoichiometry to Determine a Formula
CxHy + some oxygen ---> g CO g H2O
Now find ratio of mol H/mol C to find values of x and y in CxHy.
1.149 x mol H/ 8.61 x 10-3 mol C
= mol H / 1.00 mol C
= 4 mol H / 3 mol C
Empirical formula = C3H4
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24. Analysis of mixtures
34. At higher temperatures NaHCO3 is converted quantitatively to Na2CO3.
2 NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g)
Heating a g sample of impure NaHCO3 gives g of CO2. What was the mass percent of NaHCO3 in the original g sample?
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25. Answers 4.34 0.196 g CO2 · · · = 0.748 g NaHCO3 · 100% = 43.5% NaHCO3
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26. Analysis of mixtures
32. A g sample containing CuCl22H2O and other materials was heated. The sample mass after heating to drive off the water was g. What was the mass percent of CuCl22H2O in the original sample?
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27. Answers 4.30 (a) 10.0 g CH3SH · · · = 9.37 g CH3COSCH3
(b) · 100% = 92.3% yield
g CO2 · · · = g NaHCO · 100% = 43.5% NaHCO3
g mixture – g after heating = g H2O lost g H2O · · · = 2.41 g CuCl2·2 H2O · 100% = 91.4% CuCl2·2 H2O
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28. Stoichiometric uses – Determining a formula
44. To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen to give Fe2O3 and CO2.
If you burn g of Fex(CO)y and obtain g of Fe2O3 and g of CO2, what is the empirical formula of Fex(CO)y?
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29. Stoichiometric uses – Determining a formula
42. ■ An unknown compound has the formula CxHyOz. You burn g of the compound and isolate g of CO2 and g of H2O.
What is the empirical formula of the compound?
If the molar mass is 72.1 g/mol, what is
the molecular formula? (See Exercise 4.9.)
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30. Answers
4.44
0.799 g Fe2O3 · · = mol Fe g CO2 · · = mol CO The empirical formula is Fe(CO)5
4.42
g CO2 · · = mol C mol C · = g C g H2O · · = mol H mol H · = g H mass of O = sample mass – mass of C – mass of H = g – g C – g H = g O g O · = mol O The empirical formula is C4H8O The empirical formula mass is equal to the molar mass. The molecular formula is also C4H8O.
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31. Amounts tables and Chemical stoichiometry
18. Ethane, C2H6, burns in oxygen.
(a) What are the products of the reaction?
(b) Write the balanced equation for the reaction.
(c) What mass of O2, in grams, is required for complete
combustion of 13.6 of ethane?
(d) What is the total mass of products expected from the
combustion of 13.6 g of ethane?
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32. Answers
4.18
(a) CO2, carbon dioxide, and H2O, water (b) 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) (c) 13.6 g C2H6 · · · = 50.7 g O2 (d) 13.6 g C2H g O2 = 64.3g rxts= 64.3g products
Equation
2 C2H6(g) O2(g)  4 CO2(g) H2O(g)
I (mol) C (mol) –0.452 – E (mol)
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