Introduction to ChemistryExam III
Fall 2022
Reactions and Solutions
Name ______________________
This is an ‘At Home Exam’ due no later than Tues, Dec 6, 2022 by 11:30 p.m. on
Blackboard. Print out this exam. Neatly show all work including units and labels. Answers
with no work will only receive partial credit. If needed, use scratch paper to work out the
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pages on one page. Submit this exam as one pdf with four consecutive pages.
You can use whatever resources needed, except:
Do not copy answers from other students in the class.
Do not ask for help at the Tutoring Center.
1.
Balance the following equations:
8 pts
a.
Cr
+
b.
Al2O3
+
c.
Fe
+
O2
d.
Au2(CO3)3
+
Cl2
H2O
CrCl3
Al(OH)3
Fe2O3
NaCl
Na2CO3
+
AuCl3
2.
Classify each of the following reactions as a combination, decomposition,
4 pts single replacement, or double replacement.
Reaction Type
a. BaCl2 (aq) + K2CO3 (aq)
b. NiCl3 (l)
BaCO3 (s) + KCl (aq
Ni (s)
+
Cl2 (g)
c. AuBr (aq) + S (aq)
Au2S (aq) +
d. CO2 (aq)
ZnCO3 (s)
+
ZnO (s)
Br2 (l)
3.
For the following Oxidation/Reduction reactions, show the two half-cells and
8 pts show which element is oxidized and which is reduced.
a.
CaSO3
b.
Pb3N2
c.
Al2O3
d.
CrN
+
+
Fe
FeSO3
FeI3
PbI4
Al
+
O2
+
+
+
Ca
Fe3N2
O2
Cr2O3
+
N2
page 2
4.
8 pts
In the following reaction, sulfur (S) reacts with aluminum chloride (AlCl3) to
produce aluminum sulfide (Al2S3) and chlorine (Cl2) gas.
3 S (s)
+
2 AlCl3 (s)
1 Al2S3 (s)
+
3 Cl2 (g)
Calculate the molar mass for the following molecules.
Show all calculations and units!
4e.
a. S
b. AlCl3
c. Al2S3
d. Cl2
How many grams of AlCl3 are consumed when 2.8 moles of Al2S3 are produced?
8 pts
4f.
If 0.75 grams of S are used, how many grams of AlCl3 are needed?
8 pts
4g.
How many grams of Cl2 is 0.15 moles of Cl2?
8 pts
4h.
With 0.67 moles of AlCl3, how many grams of Al2S3 are produce?
8 pts
page 3
5.
8 pts
6.
16 pts
Calculate the concentration of a solution when a 25 mL sample of
6 M nitric acid solution is diluted to 150 mL?
A solution is prepared by combining 8 g of sodium hydroxide (NaOH)
with 245 g of H2O and the solution has a density of 1.8 g/mL.
a. What is the mass/mass percent of the sodium hydroxide solution?
b. What is the mass/volume percent of the sodium hydroxide solution?
7.
8 pts
Calculate the milliliters of solution needed to prepare a 35% (m/v) NaOH
solution containing 27 g of NaOH.
8.
How many grams of FeCl3 are generated if 125 mL of a 0.85 M BaCl2 solution
8 pts are used in this reaction?
3 BaCl2
+
1 Fe2O3
2 FeCl3
+
3 BaO
page 4
Stoichiometry Practice Quiz
Answer Key
1. What is the molar mass for each reactant and product?
a.
FeCl3
b. BaSO4
1 x Fe = 1 x 56
= 56
3 x Cl = 3 x 35.5 = 106.5
1 x Ba = 1 x 137
1 x S = 1 x 32
4 x O = 4 x 16
162.5 g FeCl3
1 mol FeCl3
c.
= 137
= 32
= 64
233 g BaSO4
1 mol BaSO4
Fe2(SO4)3
d. BaCl2
2 x Fe = 2 x 56
= 112
3 x S = 3 x 32
= 96
12 x O = 12 x 16 = 192
1 x Ba = 1 x 137 = 137
2 x Cl = 2 x 35.5 = 71
400 g Fe2(SO4)3
1 mol Fe2(SO4)3
208 g BaCl2
1 mol BaCl2
2. Balance the equation.
2 FeCl3
+
3
à
BaSO4
1 Fe2(SO4)3
+
3 BaCl2
3. How many moles of Fe2(SO4)3 are produced when 0.0045 mol of BaCl2 are produced?
1 mol Fe2(SO4)3
3 mol BaCl2
ratio
from
chemical
equation
x
0.0045 mol BaCl2
1
=
0.0015 mol Fe2(SO4)3
4. How many moles of Fe2(SO4)3 will be produced when 145 g of BaSO4 react?
1 mol Fe2(SO 4)3
1 mol BaSO4
x
3 mol BaSO4
233 g BaSO4
ratio
from
chemical
equation
145 g BaSO4
x
1
= 0.207 mol Fe2(SO 4)3
given
molar
mass
BaSO4
5. How many grams of Fe2(SO4)3 are produced when 5.6 moles of BaCl2 are produced?
400 g Fe2(SO 4)3
x
1 mol Fe2(SO 4)3
1 mol Fe2(SO 4)3
3 mol BaCl2
5.6 mol BaCl2
x
ratio
from
chemical
equation
molar
mass
Fe2(SO 4)3
= 750 g Fe2(SO 4)3
1
given
6. How many moles of FeCl3 are in 67 grams of FeCl3?
1 mol FeCl3
162.5 g FeCl3
x
67 g FeCl3
=
0.41 mol FeCl3
1
molar
mass
FeCl3
given
7. If 4.6 grams of FeCl3 are used, how many moles of BaSO4 are needed?
3 mol BaSO4
x
2 mol FeCl3
ratio
from
chemical
equation
1 mol FeCl3
162.5 g FeCl3
molar
mass
FeCl3
x
4.6 g FeCl3
1
=
0.042 mol BaSO4
given
8. In this reaction, which element is oxidized and which is reduced?
+
Cr
à
Cr2O3
+
Zn
Zn2+ O2- +
Cro
à
Cr3+ O2-
+
Zno
Zn2+
Cro
à
Cr3+ +
Zno
(remove O2- from both sides)
Zn2+
à
Zno
(gain electrons reduction)
Cro
à
Cr3+
(lose electrons oxidation)
ZnO
+
2
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