Download How to Use Reaction Stoichiometry

Chapter 4 Chemistry’s Accounting:
Reaction Stoichiometry
Quantitative information on the reaction (how much of one
reactant combines with a given amount of another).
Life as an astronaut depends
on knowing how much fuel to
load for reaching orbit and for
survival in the hostile
environment of space. These
matters of life and death
depend on calculations like
those in this chapter. We see
how to predict how much
oxygen needs to be carried to
react with a given amount of
hydrogen to power the space
shuttle and to generate
electricity. Back on Earth,
much of industry depends on
similar calculations.
How to Use Reaction Stoichiometry
Mole-to-mole calculation
2H2(g)+O2(g)2H2O(l)
1 mol O2 ~ 2 mol H2O
“chemically equivalent to”
(stoichiometric relation)
How many moles of nitrogen are needed to produce 5.0 mol of ammonia
by reaction with H2?
N2 (g)+3H2 (g)2NH3 (g)
1 mol
?
3mol
?
2 mol
5.0 mol
?= mol
Mass-to-mass calculation
How many grams of carbon dioxide are produced by burning
100.0 g of propane?
C3 H8 (g)+5O2 (g)3CO2 (g)+4H2O(l)
1 mol
3 mol
44.09 g
132 g
100.0 g
? g
? = g
Gravimetric Analysis
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Solid waste 25.40 g, silver known to be present
Concentrated nitric acid added  Ag ion
Hydrochloric acid added precipitate AgCl 16.1 g
Mass percentage of silver in the waste?
Ag  (aq)  Cl  (aq)  AgCl( s)
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1 mol
1 mol
x g
16.1 g
107.87 g
143.32 g
x=12.1 g
Mass percentage of silver = 12.1 g/25.4 g=47.7%
Case Study 4 (a)
Concentration of carbon dioxide in the atmosphere over the
past 1000 years as determined from ice core measurements.
The blue line in the inset shows the increase in the rate of
emission of CO2 since 1850.
Greenhouse gas
Case Study 4 (b)
The average surface temperature of the Earth from 1860 to
1990.
Greenhouse effect: greenhouse gases
Trap radiation and raise earth temperature
Case Study 4 (c)
The extent to which glaciers and ice caps have receded is
shown by these pictures of a boulder in the Andes. The
pictures, which were taken in 1978 (a) and 1995 (b), show
extensive local warming.
(a)
(b)
Greenhouse effect in the making!
Figure 4.6 (a) When an octane molecule undergoes complete
combustion, it forms carbon dioxide and water: one CO2
molecule is formed for each carbon atom present (yellow
arrows). (b) However, in a limited supply of oxygen, some of
the carbon atoms end up as carbon monoxide molecules, CO,
so the yield of carbon dioxide is reduced (blue arrows).
Figure 4.7
(a) The yield of a product would be 100% if no competing
reactions were taking place. (b) However, if a reactant can take
part in more than one reaction at the same time, then the yield
of a particular product will be less than 100% because other
products will also form.
(a)
(b)
The Limits of Reaction
The Limits of Reaction: An Example
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2C8H18(l)+25O2(g)16CO2(g)+18H2O(l)
1.0 L(702g)
?g
2*114.2 g
16*44.2 g
?=2160 g
(theoretical yield)
When oxygen is limited, following reaction
also occurs:
• 2C8H8(l)+17O2(g)16CO(g)+18H2O(l)
• Suppose we gathered 1.14 kg of CO2, then
• Percentage yield = 1.14/2.16=52.8%
Figure 4.8 Limiting reactant How to decide which is the
limiting reactant. (a) The gold and green boxes depict the
relative amounts of each reactant that are required by the
stoichiometric relation. (b) If the amount of reactant B is less
than that required for all A to react, then B is the limiting
reactant. (c) If the amount of A is less than that required for all B
to react, then A is the limiting reactant.
100 g of water reacts with 100.0 g of
calcium carbide. Which is limiting
reactant?
CaC2(s)+2H2O(g)Ca(OH)2(aq)+C2H2(g)
1 mol
2 mol
64.10 g 2*18.02 g=36.04 g
100 g 100 g
1.56 mol 5.55 mol
Water is too much! Only 3.12 mol is
needed!
Excess=5.55 mol –3.12 mol =2.43 mol
Combustion Analysis
After combustion:
• C CO2+NaOH(s)NaHCO3(s)
• H H2O+P4O10(s)H3PO4(l)
• Rest  total - C - H
• 1.621g newly synthesized compound
(known to contain C,H,O) after combustion
analysis yields 1.902 g of water and 3.095 g
of carbon dioxide. Determine the empirical
molecular formula.
Combustion Analysis
• 1.902 g H2O 1.902/18.02 mol H2O0.2111 mol
H
• 3.095 g CO23.095/44.01 mol CO20.07032
mol C
• Oxygen = 1.621g – 0.211*1.079g –
0.07032*12.01g = 0.564 g = 0.0352 mol
• C:H:O = 0.07032 : 0.2111 : 0.0352
= 2:6:1
• C2H6O (C2nH6nOn)
Figure 4.9 A pipet is an accurate means of transferring a fixed
volume of solution. Here, a solution containing a reactant is
being added to a reaction vessel.
Molarity 
amount of solute(moles)
volume of solution(liters)
M  Vn
1M  1 mol/L
1.345 g of potassium nitrate dissolved
In water to produce 25.0 ml of solution
1.345 g KNO3 == 1.345 g/(101.11 g/mol)
KNO3
Molarity = (1.345/101.11)mol/0.025 L =0.5321 M
Figure 4.10 The steps involved in making up a solution of
known concentration (here, a solution of potassium
permanganate, KMnO4). (a) A known mass of the compound is
dispensed into a volumetric flask. (b) Some water is added to
dissolve it. (c) Finally, water is added up to the mark. The
bottom of the solution’s meniscus, the curved top surface,
should be level with the mark.
Figure 4.11
A schematic summary of how to use molarity to convert
volume of solution to moles of solute (left) and the amount of
solute to the volume of solution that contains that amount of
solute (right).
Figure 4.12
The steps involved in dilution. A small sample of the original
solution is transferred to a volumetric flask, and then water is
added up to the mark. The diluted solution (right) has fewer
solute molecules per given volume than the concentrated
solution.
Figure 4.13
When a solution is diluted, the same number of solute
molecules occupy a larger volume, so there is a smaller
number of molecules in a given volume (indicated by the small
square).
Figure 4.14
The apparatus typically used for a titration: magnetic stirrer;
flask containing the analyte; clamp and buret containing the
titrant.
Figure 4.15
An acid-base titration in progress. The indicator is
phenolphthalein .
Figure 4.16
The schematic procedure for volume-to-volume conversions.
Figure 4.17
The schematic procedure for determining a concentration or
amount from a titration.
Oxalic acid
Titration: An Example
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• 25.0 ml of solution of oxalic acid is titrated
with 0.5 M NaOH (aq) and the
stoichiometeric point is reached when 38.0
ml of solution of base is added. Molarity of
oxalic acid?
NaOH added = 0.038 * 0.5 mol =0.019 mol
H2C2O4(aq)+2NaOHNa2C2O4(aq)+2H2O(l)
1 mol
2 mol
x mol
0.019 mol  x= 0.0095 mol
Molarity of oxalic acid = 0.0095 mol / 0.025 L
= 0.38 M
Figure 4.18
In this redox titration, the titrant is the oxidizing agent
potassium permanganate, KMnO4, and the analyte contains
iron(II) ions. The purple color of the permanganate ion
disappears as it reacts with iron(II). However, at the
stoichiometric point, the purple color persists, showing that all
the iron(II) has been oxidized to iron(III).
Connection 1 (a)
The vitamin supplements on drugstore shelves are the end
product of years of discovery, synthesis, and testing.
Connection 1 (b)
This field biologist is collecting specimens of plants that may
contain compounds with medicinal value.
Assignment for Chapter 4
19, 27, 41, 53