1. CHEMICAL REACTIONS CHAPTER 4
Reactants: Zn + I2
Product: ZnI2

2. Chapter 4 Outline Chemical Equations Stoichiometry Limiting Reactants
Chemical Analysis

3. Chemical Equations 4 Al(s) + 3 O2(g) 2Al2O3(s)
Depict the kind of reactants and products and their relative amounts in a reaction
4 Al(s) O2(g) Al2O3(s)
The numbers in the front are called stoichiometric coefficients
The letters (s), (g), and (l) are the physical states of compounds.

4. Chemical Equations 4 Al(s) + 3 O2(g) 2 Al2O3(s) This equation means
4 Al atoms + 3 O2 molecules 2 molecules of Al2O3
4 moles of Al + 3 moles of O moles of Al2O3

5. 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.
The Law of the Conservation of Matter

6. Chemical Equations Lavoisier, 1788
Because of the principle 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.
Lavoisier, 1788

7. Balancing Equations

8. Balancing Equations

9. Balancing Equations
___C3H8 (g) + ___ O2 (g) > ___CO2 (g) ___ H2O (g)
C3H8 (g) O2 (g) > CO2 (g) H2O (g)

10. Balancing Equations
___B4H10 (g) + ___ O2 (g) ---> ___ B2O3 (g) + ___ H2O (g)
2 B4H10 (g) O2 (g) ---> B2O3 (g) H2O (g)

11. 11
STOICHIOMETRY
- the study of the quantitative aspects of chemical reactions.

12. PROBLEM: If 454 g of NH4NO3 decomposes, how much H2O and N2O are formed? What is the theoretical yield of products?
STEP 1
Write the balanced chemical equation
NH4NO N2O H2O

13. Convert reactant mass to moles
454 g of NH4NO3 --> N2O + 2 H2O
STEP 2
Convert reactant mass to moles
(454 g) --> moles
454 g •
1 mol
80.04 g =
5.68 mol NH 4 NO 3

14. 2 mol H O produced 1 mol NH NO used 4 3 STEP 3
454 g of NH4NO3 --> N2O + 2 H2O
STEP 3
Convert moles reactant --> moles product.
Relate moles NH4NO3 to moles product.
1 mol NH4NO3 --> 2 mol H2O
Express this relation as the
STOICHIOMETRIC FACTOR 2
mol H
O produced
1 mol NH 4 NO 3
used

15. Convert moles reactant (5.68 mol) moles product
454 g of NH4NO3 --> N2O + 2 H2O
STEP 3
Convert moles reactant (5.68 mol) moles product
5.68 mol NH 4 NO 3 • 2
mol H
O produced
1 mol NH
used
= mol H2O produced

16. STEP 4 454 g of NH4NO3 --> N2O + 2 H2O THEORETICAL YIELD
Convert moles product (11.4 mol) to mass product.
This is called the
THEORETICAL YIELD
This is the “Expected” # of moles.

17. STEP 4 454 g of NH4NO3 --> N2O + 2 H2O
Convert moles prod. (11.4 mol) to mass prod. 18 . 02 g
11.4 mol H O • =
204 g H O 2 2
1 mol
This is the “Expected” Mass!
ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!
Repeat to find the grams of N2O formed.

18. STEP 5 454 g of NH4NO3 --> N2O + 2 H2O How much N2O is formed?
Total mass of reactants=total mass of products
454 g NH4NO3 = ___ g N2O g H2O
mass of N2O = 250. g
This is an alternate method.

19. STEP 6 454 g of NH4NO3 --> N2O + 2 H2O Calculate the percent yield.
If you isolated only 131 g of N2O, what is the percent yield?
This compares the theoretical (250. g) and actual (131 g) yields.

20. Calculate the percent yield.
454 g of NH4NO3 --> N2O + 2 H2O
STEP 6
Calculate the percent yield. %
yield =
actual yield
theoretical yield •
100% %
yield =
131 g
250. g •
100%
52.4%

21. Stoichiometry Calculations
General Plan For
Stoichiometry Calculations
Mass
reactant
Stoichiometric
factor
Moles
product

22. 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Reaction is catalyzed by MnO2
PROBLEM: Using 5.00 g of H2O2, what mass of O2 and of H2O can be obtained?
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Reaction is catalyzed by MnO2
Step 1: moles of H2O2
Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O2
Step 3: mass of O2
Repeat for H2O.

23. Reactions Involving a LIMITING REACTANT
In a given reaction, there is not enough of one reagent to use up the other reagent completely.
The reagent in short supply LIMITS the quantity of product that can be formed.

24. LIMITING REACTANTS 2 NO(g) + O2 (g) 2 NO2(g) NO
Products
2 NO(g) + O2 (g) NO2(g)
Limiting reactant = ___________ Excess reactant = ____________ NO O2

25. LIMITING REACTANTS

26. LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq)
Zn + 2 HCl ---> ZnCl2 + H2
Rxn Rxn Rxn 3
mass Zn (g)
mol Zn
mol HCl
mol HCl/mol Zn

27. PROBLEM: Mix 5. 40 g of Al with 8. 10 g of Cl2
PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. How many grams of Al2Cl6 can form?

28. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

29. mol Cl mol Al = 3 2 Al + 3 Cl2 ---> Al2Cl6 2
Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.
2 Al Cl2 ---> Al2Cl6
Reactants must be in the mole ratio
mol Cl 2
mol Al = 3

30. Deciding on the Limiting Reactant
2 Al Cl2 Al2Cl6
mol Cl 2
mol Al
> 3 If
then there is not enough Al to use up all the Cl2, and the limiting reagent is
Al.

31. Deciding on the Limiting Reactant
2 Al Cl2 ---> Al2Cl6
mol Cl 2
mol Al
< 3 If
then there is not enough Cl2 to use up all the Al, and the limiting reagent is
Cl2

32. Step 2 of LR problem: Calculate moles of each reactant
We have 5.40 g of Al and 8.10 g of Cl2 5 . 40
g Al •
1 mol
27.0 g =
0.200 mol Al
1 mol 8
.10 g Cl • =
0.114 mol Cl 2 2
70.9 g

33. Find mole ratio of reactantsCl 2
mol Al =
0.114 mol
0.200 mol
0.57
This
should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio.
Limiting reagent is Cl2

34. Base all calculations on Cl2
Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
Limiting reactant = Cl2
Base all calculations on Cl2
moles
Cl2
Al2Cl6
grams

35. CALCULATIONS: calculate mass of Al2Cl6 expected.
Step 1: Calculate moles of Al2Cl6 expected based on LR. .
114
mol Cl 2 • 1 Al 6
3 mol Cl =
mol Al
Step 2: Calculate mass of Al2Cl6 expected based on LR. .
0380
mol Al 2 Cl 6 •
66.4 g Al =
10.1 g Al

36. How much of which reactant will remain when reaction is complete?
Cl2 was the limiting reactant.
Therefore, Al was present in excess.
But how much?
First find how much Al was required.
Then find how much Al is in excess.

37. Calculating Excess Al
2 Al Cl2
products
0.114 mol = LR
0.200 mol

38. Calculating Excess Al 2 Al + 3 Cl2 0.114 mol = LR
products
0.114 mol = LR
0.200 mol
2 mol Al
.114 mol Cl • =
mol Al req' d 2
3 mol Cl 2
Excess Al = Al available - Al required
= mol mol
= mol Al in excess

39. Nitrogen and iodine react to form nitrogen tri iodide. If 50
Nitrogen and iodine react to form nitrogen tri iodide. If 50.0 g of nitrogen is mixed with g iodine, calculate the number of grams of product formed and the grams of reactant remaining.
N I NI3
28.0 g/mol g/mol g/mol
1.79 mole mole
L.R. (0.460 S.U.)
50.0g -12.9g g left g
37.1g left g g = 400.0g = 363g g

40. Using Stoichiometry to Determine a Formula
Burn g of a hydrocarbon, CxHy, and produce g of CO2 and g of H2O.
What is the empirical formula of CxHy?
CxHy + oxygen g CO g H2O 40

41. Chemical Molecular Analysis
in the lab

42. 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 moles of C in CO2
8.61 x 10-3 mol C
2. Calculate moles of H in H2O
1.149 x mol H

43. 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

44. Chemical Analysis Combustion-Determine a Formula
CxHy + some oxygen g CO g H2O
Recognize that all C in CO2 and all H in H2O is from CxHy.
1. Calculate moles of C in CO2
0.379 g CO2 mole CO mole C
44.0 g CO2 mole CO2
= mole C

45. Chemical Analysis Combustion-Determine a Formula
CxHy+some oxygen gCO g H2O
2. Calculate moles of H in H2O
g H2O mole H2O 2 mole H
18.0 g H2O mole H2O
= mole H

46. Chemical Analysis Combustion-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.
mol H/ mol C
= mol H / 1.00 mol C
= mol H / 3.00 mol C
Empirical formula = C3H4

47. Sample Problems
1) A g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced g of CO2 and g of H2O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula.
1.030 g CO2 mole CO mole C g C g CO mole CO2 mole C
= g C
0.632 g H2O mole H2O 2 mole H g H g H2O mole H2O mole H
= g H

48. Sample Problems 0.537 g C, H, O - 0.281 g C - 0.070 g H 0.186 g O
1) A g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced g of CO2 and g of H2O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula.
0.537 g C, H, O
g C
g H
0.186 g O

49. Sample Problems C H O .281 g 1 mole 12.0 g .070 g 1 mole 1.0 g .186 g
Empirical formula C2H6O

50. Sample Problems
Empirical formula C2H6O 90 46
= 2
Molecular formula C4H12O2
Alternate method. Use grams of cmpd. and its molar mass to determine moles of cmpd. Divide moles of C and H by these moles to find the subscripts for C and H. The subscript for O can be determinded by difference.

51. Sample Problems
2) A g sample of ascorbic acid, vitamin C, was burned to produce g of CO2 and g H2O. Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula.
Given that the molecular weight is approximately 180 g/mole, determine the molecular formula.
.1869 g CO2 mole CO mole C g C g CO mole CO2 mole C
= g C
.0510 g H2O mole H2O mole H g H g H2O mole H2O mole H
= g H

52. Sample Problems 0.1247 g C, H, O - 0.0510 g C - 0.0057 g H 0.0680 g O
2) A g sample of ascorbic acid, vitamin C, was burned to produce g of CO2 and g H2O. Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula. Given that the molecular weight is approximately 180 g/mole, determine the molecular formula.
g C, H, O
g C
g H
g O

53. Sample Problems C H O .0510g 1 mole 12.0 g .0057g 1 mole 1.0 g .0680g
Empirical formula C3H4O3

54. Sample Problems Empirical formula C3H4O3 180 = 2 88
Molecular formula C6H8O6

55. Chemical Analysis Mixtures-Determine a Percent
1. The amount of calcium present in milk can be determined by adding oxalate ion, C2O42-(in the form of its water-soluble sodium salt, Na2C2O4); the insoluble compound calcium oxalate is precipitated. Suppose you take a 75.0 g sample of milk and isolate g of calcium oxalate from it. What is the weight percentage of calcium in the milk?
MILK (Ca2+)
Na2C2O4
CaC2O4

56. Chemical Analysis Mixtures-Determine a Percent
MILK (Ca2+)
Na2C2O4
CaC2O4
75.0 g g
? % Ca
% Ca =
g Ca
75.0 g milk
X 100

57. Chemical Analysis Mixtures-Determine a Percent
MILK (Ca2+)
Na2C2O4
CaC2O4
75.0 g g
? % Ca
.288 g CaC2O4 mole CaC2O mole Ca g Ca
128.1 g CaC2O4 mole CaC2O mole Ca
= g Ca
% Ca =
g Ca
75.0 g milk
X 100
= 0.120%

58. Chemical Analysis Mixtures-Determine a Percent
2. A 4.22 g sample of calcium chloride and sodium chloride was dissolved in water, and the solution was treated with sodium carbonate to precipitate the calcium as calcium carbonate. After isolating the solid calcium carbonate, it was heated to drive off the carbon dioxide and form g of calcium oxide. What is the weight percent of calcium chloride in the original 4.22 g sample?

59. Chemical Analysis Mixtures-Determine a Percent
CaCl2/NaCl
Na2CO3
CaCO3
4.22 g
Heat
% CaCl2 =
g CaCl2
4.22 g sample
X 100
CaO
0.959 g

60. Chemical Analysis Mixtures-Determine a Percent
0.959 g CaO mole CaO mole CaCl g CaCl2
56.1 g CaO mole CaO mole CaCl2
= g CaCl2
% CaCl2 =
1.90 g CaCl2
4.22 g sample
X 100
= 45.0%

61. Practice Problems 1. Balance the following equations:
CS2 + Cl2 --> CCl4 + S2Cl2
N2 + O2 --> NO
C8H O2 --> CO2 + H2O
2. Write the formula equation for each of the following:
sodium + water --> sodium hydroxide + hydrogen
magnesium + oxygen --> magnesium oxide
aluminum + hydrochloric acid -->
aluminum chloride + hydrogen

62. Practice Problems 2. (continue) aluminum + hydrochloric acid -->
aluminum chloride + hydrogen
sodium chlorate --> sodium chloride + oxygen
mercury(II) sulfate + ammonium sulfide -->
mercury(II) sulfide + ammonium sulfate
iron + cupric sulfate --> iron(III) sulfate + copper

63. Practice Problems For problems 3-6 3 H2 + N2 g 2 NH3
3. How many moles of H2 are required to react 4.2 moles of N2?
4. How many moles of H2 are required to react 74 grams of N2?
5. How many grams of NH3 would be produced from 45 g of H2?
6. How many moles of NH3 would be produced from the reaction of 18.5 g H2 and 95 g of N2?

64. Practice Problems
7. Phosphine, PH3, is formed when calcium phosphide is added to water. How many grams of phosphine can be obtained from 205 g of calcium phosphide?
8. How many grams of iron will be required to release all of the antimony from 10.0 g antimony trisulfide? (Ferrous sulfide is formed)
9. If calcium oxide were prepared by heating calcium carbonate, how many grams of the carbonate would be required to produce the 15.0 g of the oxide?

65. Practice Problems
10. How many grams of cupric sulfate are needed to completely react with 145 g of sodium chloride? How many grams of sodium sulfate could be produced by this reaction?
11. How many grams of sulfuric acid will react with 40.0 g of aluminum metal?
grams of zinc are added to 120 grams of nitrous acid. How many grams of hydrogen gas are evolved?

66. Practice Problems
grams of hydrogen and 75 grams of oxygen are exploded together in a reaction tube. How many grams of water are produced? What other gas is found in the tube(besides water vapor) after the reaction, and how many grams of this gas are there?

67. Practice Problems
14. A white powder was a mixture of NaBr and NaNO3. A sample of the powder weighing g was dissolved in water and a solution of AgNO3 was added until the precipitation of AgBr was complete. The reaction mixture was filtered and dried and the precipitate of AgBr weighed g. What was the percentage by weight of NaBr in the original sample?

68. Practice Problems
15. A 4.81 g sample of an unknown compound containing only carbon, hydrogen, and nitrogen was burned to produce g of CO2 and 1.68 g of H2O. Determine the empirical formula.

69. Practice Problems Answers
1. 1,3,1,1 1,1,2 2,25,16,18
2. 2 Na + 2 HOH --> 2 NaOH + H2
2 Mg + O2 --> 2 MgO
2 Al + 6 HCl --> 2 AlCl3 + 3 H2
2 NaClO3 --> 2 NaCl + 3 O2
HgSO4 + (NH4)2S --> Hg + (NH4)2SO4
2 Fe + 3 CuSO4 --> Fe2(SO4)3 + 3 Cu
3. 13 mole H mole H2
g NH mole NH3

70. Practice Problems Answers
g PH3
g Fe
g CaCO3
g CuSO4, 176 g Na2SO4
g H2SO4
g H2
g H2O, 0.6 g H2 %
15. C5H3N

71. Writing Equations zinc + chlorine ---> zinc chloride
Zn (s) Cl2 (g) --> ZnCl2 (s)
Combination, Synthesis

72. Writing Equations KNO3 (s) --> KNO2 (s) + O2 (g)
potassium nitrate --> potassium nitrite + oxygen
KNO3 (s) --> KNO2 (s) O2 (g)
2 KNO3 (s) --> 2 KNO2 (s) O2 (g)
Decomposition

73. Writing Equations Single Displacement
magnesium bromide + chlorine --> magnesium chloride + bromine
MgBr2 (s) Cl2 (g) --> MgCl2 (s) + Br2 (g)
Single Displacement

74. Writing Equations Double Displacement
calcium hydroxide + hydrochloric acid --> calcium chloride + water
Ca(OH)2 (aq) + HCl (aq) --> CaCl2 (aq) + H2O (l)
Ca(OH)2 (aq) + 2HCl (aq) --> CaCl2 (aq) H2O (l)
Double Displacement

75. Writing Equations
zinc chloride + ammonium sulfide --> zinc sulfide + ammonium chloride
ZnCl2 (aq) + (NH4)2S (aq) --> ZnS (s) + NH4Cl (aq)
ZnCl2 (aq) + (NH4)2S (aq) --> ZnS (s) NH4Cl (aq)

76. Writing Equations
aluminum + cupric chloride --> copper + aluminum chloride
Al (s) + CuCl2 (aq) --> Cu (s) + AlCl3 (aq)
2 Al (s) + 3 CuCl2 (aq) --> 3 Cu (s) + 2 AlCl3 (aq)

77. PROBLEM: How many moles of H2 are required to produce 9 moles of NH3?
STEP 1
Write the balanced chemical equation
3 H N > 2 NH3

78. STEP 2 Write the given and requested information below the equation.
3 H N > 2 NH3
? mole mole

79. STEP 3 9 mole NH3 3 mole H2 2 mole NH3 = 10 mole H2
Calculate using the information.
3 H N > 2 NH3 ? mole mole
9 mole NH mole H2
2 mole NH3
= 10 mole H2

80. PROBLEM: How many moles of NH3 can be produced from 10.4 moles of N2?
3 H N > 2 NH mole ? mole
10.4 mole N mole NH3
mole N2
= 20.8 mole NH3

81. PROBLEM: How many grams of H2 are required to produce 8 moles of NH3?
3 H N > 2 NH3 ? g 8 mole
8 mole NH3 3 mole H g H2
2 mole NH3 mole H2
= 20 g H2

82. PROBLEM: How many moles of NH3 can be produced from 55 grams of N2?
3 H N > 2 NH g ? mole
55 g N mole N mole NH3
28.0 g N mole N2
= 3.9 mole NH3

83. PROBLEM: How many grams of H2 are required to react 24 grams of N2?
3 H N > 2 NH3 ? g g
24 g N2 mole N mole H g H2
28.0 g N2 mole N mole H2
= g H2

84. PROBLEM: How many grams of N2 are required to produce 155 grams of NH3?
3 H N > 2 NH ? g g
155 g NH3 mole NH3 mole N g N2
17.0 g NH3 2 mole NH3 mole N2
= 128 g N2

85. Sample Problems
1) Sulfur dioxide may be oxidized to sulfur trioxide. How many grams of sulfur dioxide could be converted by this process if g of oxygen are available for the oxidation?
2 SO O > 2 SO3
? g g
100.0 g O2 mole O mole SO g SO2
32.0 g O mole O mole SO2
= 401 g SO2

86. Sample Problems
2) Lightning discharges in the atmosphere catalyze the conversion of nitrogen to nitrogen dioxide. How many grams of nitrogen would be required to make 25.0 g of nitrogen dioxide in this way?
N O > 2 NO2
? g g
25.0 g NO2 mole NO mole N g N2
46.0 g NO mole NO2 mole N2
= g N2

87. Sample Problems
3) Ferric oxide may be reduced to pure iron with coke (pure carbon). Suppose that g of ferric oxide is available. How many grams of carbon would be needed?
2 Fe2O C Fe CO g
? g
150.0 g Fe2O3 mole Fe2O mole C g C
159.6 g Fe2O3 2 mole Fe2O3 mole C
= g C

88. Sample Problems
4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas would be produced?
Zn HCl > ZnCl H2
50.0 g ? g ? g
50.0 g Zn mole Zn 2 mole HCl 36.5 g HCl
65.4 g Zn mole Zn mole HCl
= 55.8gHCl

89. Sample Problems 50.0 g Zn mole Zn mole H2 2.0 g H2
4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas would be produced?
Zn HCl > ZnCl H2
50.0 g ? g ? g
50.0 g Zn mole Zn mole H g H2
65.4 g Zn mole Zn mole H2
= 1.5 g H2

90. Sample Problems
5) Phosphoric acid, H3PO4, is produced in the reaction between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be produced from 55 grams of calcium phosphate? What other product is formed and in what quantity?
Ca3(PO4) H2SO4 --> 3 CaSO H3PO g
? g ? g
55 g Ca3(PO4)2 mole Ca3(PO4) mole CaSO g CaSO4
310.3 gCa3(PO4)2 mole Ca3(PO4)2 mole CaSO4
= 72 g CaSO4

91. Sample Problems
5) Phosphoric acid, H3PO4, is produced in the reaction between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be produced from 55 grams of calcium phosphate? What other product is formed and in what quantity?
Ca3(PO4) H2SO4 --> 3 CaSO H3PO g
? g ? g
55 g Ca3(PO4)2 mole Ca3(PO4) mole H3PO g H3PO4
310.3 gCa3(PO4)2 mole Ca3(PO4)2 mole H3PO4
= 35 g H3PO4

92. Percent Yield %
yield =
actual yield
theoretical yield •
100%

93. PROBLEM: If 19.3 g H2 produces 78.5 g NH3, what is the percent yield?
3 H N > 2 NH g ? g
19.3 g H2 mole H2 2 mole NH g NH3
2.0 g H2 3 mole H mole NH3
= 110 g NH3
78.5g
110 g
% yield =
x = 71 %

94. PROBLEM: If the yield obtained is 75%, how many grams of NH3 would be obtained from 10.4 g of N2?
3 H N > 2 NH3
10.4 g ? g
10.4 g N2 mole N mole NH g NH3
28.0 g N2 mole N mole NH3
= 12.6 g NH3
12.6 g NH3 x .75 = 9.4 g NH3

95. PROBLEM: grams of potassium hydroxide are permitted to react with 50.0 grams of hydrochloric acid. How many grams of potassium chloride are formed?
STEP 1
Write the balanced chemical equation
KOH HCl --> KCl + HOH

96. STEP 2 Write the given and requested information below the equation.
KOH HCl --> KCl + HOH
75.0 g g ?g

97. STEP 3 75.0 g KOH mole KOH mole KCl 74.6 g KCl
Calculate the product assuming that each reactant is the limiting reagent.
KOH HCl --> KCl + HOH
75.0 g g ?g
75.0 g KOH mole KOH mole KCl g KCl
56.1 g KOH mole KOH mole KCl
= 99.7g KCl
50.0 g HCl mole HCl mole KCl g KCl
36.5 g HCl mole HCl mole KCl
= 102 g KCl

98. STEP 4 75.0 g KOH can produce 99.7 g KCl
Determine the limiting reactant and the actual amount of product.
KOH HCl --> KCl + HOH
75.0 g g ?g
75.0 g KOH can produce 99.7 g KCl
KOH is the limiting reactant, HCl is the excess reactant.
50.0 g HCl can produce 102 g KCl
99.7 g KCl

99. STEP 5 75.0 g KOH mole KOH mole HCl 36.5 g HCl
Determine the amount of excess reactant by calculating the amount used and subtracting from the starting amount.
KOH HCl --> KCl + HOH
75.0 g g ?g
75.0 g KOH mole KOH mole HCl g HCl
56.1 g KOH mole KOH mole HCl
= 48.8g HCl
used
50.0 g = 1.2 g HCl left

100. Sample Problems
1) How many grams of carbon dioxide can be obtained from the action of grams of hydrobromic acid on grams of calcium carbonate?
2 HBr + CaCO > HOH + CO2 + CaBr2
100.0 g g ? g
100.0 g HBr mole HBr mole CO g CO2
80.9 g HBr 2 mole HBr mole CO2
= 27.2 g CO2
100.0 g CaCO3 mole CaCO mole CO g CO2
100.1 g CaCO3 mole CaCO3 mole CO2
= 44.0 g CO2

101. Sample Problems
2) How many grams of ammonia are evolved when 34 grams of ammonium chloride are added to 140 grams of barium hydroxide?
2 NH4Cl + Ba(OH) > BaCl NH HOH
34 g g ? g
34 g NH4Cl mole NH4Cl mole NH g NH3
53.5 g NH4Cl 2 mole NH4Cl mole NH3
= 11 g NH3
140 g Ba(OH)2 mole Ba(OH) mole NH g NH3
171.3 g Ba(OH)2 mole Ba(OH)2 mole NH3
= 28.0 gNH3

102. Sample Problems
3) grams of lithium metal is dropped into liter of water. How many grams of hydrogen are produced?
2 Li HOH > 2 LiOH H2
100.0 g g ? g
100.0 g Li mole Li mole H g H g Li mole Li mole H2
= 14 g H2
1000. g HOH mole HOH mole H g H g HOH mole HOH mole H2
= 56 g H2

103. Sample Problems 50.0 g O2 mole O2 mole C 12.0 g C
4) grams of oxygen are available for the combustion of 25.0 grams of carbon. How many grams in excess is the oxygen or carbon ?
O C > CO2
? g excess g g
50.0 g O2 mole O mole C g C
32.0 g O mole O2 mole C
= 18.8g C used
25.0 g = g C left

104. Sample Problems
5) grams of sulfuric acid is added to grams of barium peroxide. Which reactant is in excess and by how many grams?
H2SO BaO > BaSO H2O2
g excess g g
140.gH2SO4 mole H2SO4 mole BaO g BaO g H2SO4 mole H2SO4 mole BaO2
= g BaO2
Since only 230. g available, BaO2 is the limiting reactant

105. Sample Problems = 133 g H2SO4 used 140. g - 133 = 7 g H2SO4 left
5) grams of sulfuric acid is added to grams of barium peroxide. Which reactant is in excess and by how many grams?
H2SO BaO > BaSO H2O2
g excess g g
230. g BaO2 mole BaO mole H2SO g H2SO g BaO2 mole BaO2 mole H2SO4
= g H2SO4 used
140. g = 7 g H2SO4 left