Practice Problems
1.
Potassium nitrate decomposes on heating,
producing potassium oxide and gaseous
nitrogen and oxygen. To produce 56.6 kg
of oxygen, how many (a) moles of KNO3
must be heated? (b) grams of KNO3 must
be heated?
4 July 2013
1
1.
Potassium nitrate decomposes on heating,
producing potassium oxide and gaseous
nitrogen and oxygen. To produce 56.6 kg of
oxygen, how many (a) moles of KNO3 must be
heated? (b) grams of KNO3 must be heated?
4KNO3(s)  2K2O(s) + 2N2(g) + 5O2(g)
𝑚𝑜𝑙 𝐾𝑁𝑂3 = 56.6 𝑘𝑔 𝑂2
1000 𝑔
1 𝑘𝑔
1000 𝑔
1 𝑘𝑔
= 1.43𝑥105 𝑔 𝐾𝑁𝑂3
𝑔 𝐾𝑁𝑂3 = 56.6 𝑘𝑔 𝑂2
1 𝑚𝑜𝑙 𝑂2
32.00 𝑔 𝑂2
1 𝑚𝑜𝑙 𝑂2
32.00 𝑔 𝑂2
4 𝑚𝑜𝑙 𝐾𝑁𝑂3
= 1.42x103 mol KNO3
5 𝑚𝑜𝑙 𝑂2
4 𝑚𝑜𝑙 𝐾𝑁𝑂3
5 𝑚𝑜𝑙 𝑂2
4 July 2013
101.11 𝑔 𝐾𝑁𝑂3
1 𝑚𝑜𝑙 𝐾𝑁𝑂3
2
Practice Problems
2.
Chromium(III) oxide reacts with hydrogen
sulfide (H2S) gas to form chromium(III) sulfide
and water. To produce 421 g of Cr2S3, (a) how
many moles of Cr2O3 are required? (b) grams
of Cr2O3 are required?
4 July 2013
3
2.
Chromium(III) oxide reacts with hydrogen
sulfide (H2S) gas to form chromium(III) sulfide
and water. To produce 421 g of Cr2S3, (a) how
many moles of Cr2O3 are required? (b) grams
of Cr2O3 are required?
Cr2O3(s) + 3H2S(g)  Cr2S3(s) + 3H2O(l)
𝑚𝑜𝑙 𝐶𝑟2𝑂3 = 421 𝑔 𝐶𝑟2𝑂3
= 2.10 𝑚𝑜𝑙 𝐶𝑟2𝑂3
1 𝑚𝑜𝑙 𝐶𝑟2𝑆3
200.21 𝑔 𝐶𝑟2𝑆3
1 𝑚𝑜𝑙 𝐶𝑟2𝑆3
200.21 𝑔 𝐶𝑟2𝑆3
= 3.20𝑥102 𝑔 𝐶𝑟2𝑂3
𝑔 𝐶𝑟2𝑂3 = 421 𝑔 𝐶𝑟2𝑂3
1 𝑚𝑜𝑙 𝐶𝑟2𝑂3
1 𝑚𝑜𝑙 𝐶𝑟2𝑆3
1 𝑚𝑜𝑙 𝐶𝑟2𝑂3
1 𝑚𝑜𝑙 𝐶𝑟2𝑆3
4 July 2013
152.00 𝑔 𝐶𝑟2𝑂3
1 𝑚𝑜𝑙 𝐶𝑟2𝑂3
4
Practice Problems
3.
Calculate the mass of each product
formed when 43.82 g of diborane (B2H6)
reacts with excess water:
Unbalanced equation:
B2H6(g) + H2O(l)  H3BO3(s) + H2(g)
4 July 2013
5
3.
Calculate the mass of each product formed
when 43.82 g of diborane (B2H6) reacts with
excess water:
B2H6(g) + 6H2O(l)  2H3BO3(s) + 6H2(g)
𝑔 𝐻3𝐵𝑂3 = 43.82 𝑔 𝐵2𝐻6
= 195.8 𝑔
𝑔 𝐻2 = 43.82 𝑔 𝐵2𝐻6
= 19.16 𝑔
1 𝑚𝑜𝑙 𝐵2𝐻6
27.67 𝑔 𝐵2𝐻6
1 𝑚𝑜𝑙 𝐵2𝐻6
27.67 𝑔 𝐵2𝐻6
2 𝑚𝑜𝑙 𝐻3𝐵𝑂3
1 𝑚𝑜𝑙 𝐵2𝐻6
6 𝑚𝑜𝑙 𝐻2
1 𝑚𝑜𝑙 𝐵2𝐻6
4 July 2013
61.83 𝑔 𝐻3𝐵𝑂3
1 𝑚𝑜𝑙 𝐻3𝐵𝑂3
2.016 𝑔 𝐻2
1 𝑚𝑜𝑙 𝐻2
6
Practice Problems
4.
Calculate the mass of each product
formed when 174 g of silver sulfide reacts
with excess hydrochloric acid:
Unbalanced equation:
Ag2S(s) + HCl(aq)  AgCl(s) + H2S(g)
4 July 2013
7
4.
Calculate the mass of each product formed
when 174 g of silver sulfide reacts with excess
hydrochloric acid:
Ag2S(s) + 2 HCl(aq)  2 AgCl(s) + H2S(g)
𝑔 𝐴𝑔𝐶𝑙 = 174 𝑔 𝐴𝑔2𝑆
= 201 𝑔 𝐴𝑔𝐶𝑙
𝑔 𝐻2𝑆 = 174 𝑔 𝐴𝑔2𝑆
= 23.9 𝑔 𝐻2𝑆
1 𝑚𝑜𝑙 𝐴𝑔2𝑆
247.9 𝑔 𝐴𝑔2𝑆
2 𝑚𝑜𝑙 𝐴𝑔𝐶𝑙
1 𝑚𝑜𝑙 𝐴𝑔2𝑆
143.4 𝑔 𝐴𝑔𝐶𝑙
1 𝑚𝑜𝑙 𝐴𝑔𝐶𝑙
1 𝑚𝑜𝑙 𝐴𝑔2𝑆
247.9 𝑔 𝐴𝑔2𝑆
1 𝑚𝑜𝑙 𝐻2𝑆
1 𝑚𝑜𝑙 𝐴𝑔2𝑆
34.09 𝑔 𝐻2𝑆
1 𝑚𝑜𝑙 𝐻2𝑆
4 July 2013
8
Limiting Reagent

The reactant that will run out first (is
consumed) and determines the quantity
or amount of product that can be formed
4 July 2013
9
Which is the limiting reagent?

If 25.0 g CH4 is combusted with 40.0 g O2, which
reactant is the limiting reagent?
There are 2 ways to identify the limiting reagent:
METHOD 1. For both reactants, use the balanced
equation & stoichiometric factors to compute the
amount of any product formed.
METHOD 2. Pick one of the reactants and compute
how much of the other reactant you need.
Compare with how much is actually available.
4 July 2013
10

If 25.0 g CH4 is combusted with 40.0 g O2, which
reactant is the limiting reagent?
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
METHOD 1. Calculate the # of mol product formed by each
reactant to determine which reactant makes the least amount.
Let’s use the amount of CO2 formed as our “yardstick.” (We
could also choose H2O.)
Since O2 produced the lesser amount of product, it must be
the limiting reagent.
4 July 2013
11

If 25.0 g CH4 is combusted with 40.0 g O2, which
reactant is the limiting reagent?
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
METHOD 2. Directly compare amounts of reactants given in
the problem.
How many grams of O2 is needed to completely react with
25.0 g CH4?
O2 is the limiting reagent since we need 99.75 g of it, but we
are only given 40.0 g O2. Thus, the amount of product that can
be formed is determined by the amount of O2 and not by the
amount of methane.
4 July 2013
12
Practice Exercise
Metal hydrides react with water to form hydrogen
gas and the metal hydroxide. For example,
SrH2(s) + 2H2O(l)  Sr(OH)2(s) + 2H2(g)
You wish to calculate the mass of H2 that can be
prepared from 5.70 g of SrH2 and 4.75 g of H2O.
a. How many moles of H2 can form from the given
mass of SrH2?
b. How many moles of H2 can form from the given
mass of H2O?
c. Which is the limiting reactant?
d. How many grams of H2 can form?
Yields
1.
2.
3.
Theoretical yield: amount of product
calculated using the molar ratios from
the balanced equation
Actual yield: amount of product actually
obtained
Percent yield
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
% 𝑦𝑖𝑒𝑙𝑑 =
× 100
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
4 July 2013
14
Practice Problem

Silicon carbide (SiC) is made by reacting
sand (silicon dioxide, SiO2) with
powdered carbon at high temperature.
Carbon monoxide is also formed. What
is the percent yield if 51.4 kg of SiC is
recovered from processing 100.0 kg of
sand? (MW of SiC: 40.10 g/mol; SiO2:
60.09 g/mol)
4 July 2013
15

Silicon carbide (SiC) is made by reacting sand (silicon
dioxide, SiO2) with powdered carbon at high
temperature. Carbon monoxide is also formed. What is
the percent yield if 51.4 kg of SiC is recovered from
processing 100.0 kg of sand?
SiO2(s) + 3C(s) → SiC(s) + 2CO(g)
3
100.0 kg SiO2 x 10 g x 1 mol SiO2 = 1664 mol SiO
2
1 kg
60.09 g SiO2
mol SiO2 = mol SiC = 1664 mol SiC
1664 mol SiC x 40.10 g SiC x 1 kg = 66.73 kg
1 mol SiC
103g
51.4 kg
66.73 kg
x 100
= 77.0%
4 July 2013
16
Solution Stoichiometry

A solution is a homogenous mixture of a
solvent plus a solute.
4 July 2013
17
Other Solutions
4 July 2013
18
Properties of a Solution

Distribution of solute in solvent
is uniform

Components do not separate or
sediment on standing

Not separable by filtration

Solute/solvent mix in ratios up
to the solubility limit of the
solute
4 July 2013
19
Concentration of Solutions
Amount of solute present in a given
quantity of solvent or solution
 Expressed as molarity

𝑔𝑟𝑎𝑚𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑀 = 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
𝑡𝑜𝑡𝑎𝑙 𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
4 July 2013
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
20
Concentration

An intensive (not extensive) quantity
◦ Independent of the volume of solution (like
density or temperature)
◦ Example: a 50 L tank of a solution has the
same concentration as a 50 mL beaker of the
same solution
4 July 2013
21
4 July 2013
22
Practice Problem

Calculate the molarity of a solution that
contains 12.5 g of pure sulfuric acid
(H2SO4) in 1.75 L of solution (Molar mass
H2SO4 = 98.1 g)
1 𝑚𝑜𝑙 𝐻2𝑆𝑂4
12.5 𝑔 𝐻2𝑆𝑂4 98.1 𝑔 𝐻 𝑆𝑂
2
4
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
= 0.0728 M 𝐻2𝑆𝑂4
1.75 𝐿
4 July 2013
23
Preparing Solutions in the
Laboratory
4 July 2013
24
Problem Solving Classes
Ch 7 D
F 12:30 – 1:30
SOM 202
Ch 7 E
Ch 7 F
Ch 7 G
Ch 7 H-K
Ch 7 L
Ch 7 C
F 11:30 – 12:30
M 12:30 – 1:30
W 2:30 – 3:30
MWF 3:30 – 4:30
T 4:30 – 5:30
TH 4:30 – 5:30
CTC 102
CTC 104
SOM 105
C 109
C 114
C 114
4 July 2013
25
Problem Sets
Chapter
1
2
3
Practice problems
6, 7, 8, 22, 24, 26, 30, 34, 47, 49, 53, 55
11, 13, 17, 33, 61, 65, 67, 69, 71, 73, 75
6, 10, 12, 14, 16, 18, 20, 23, 26, 28, 30, 32,
33, 38, 40, 41, 43, 45, 49, 51, 65, 66, 69, 71,
73, 75, 83, 85, 89, 93
4 July 2013
26