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Which lab section are you in?
(Circle One)
EXAM 4 REVIEW
CHEM 110

M
W
R
Note: As Part 1 is supposed to be memorized material, you can cross-check your answers
with the information in your textbook and/or from your notes.
PART 2: MATH
15a)
Calculate the molar mass of Na2Cr2O7. Rewrite final answer 
261.98 g/mol Na2Cr2O7
(Time Range: 3 to 5 min)
g/mol
g/mol
2 × Na

2×
22.99

45.98
2 × Cr

2×
52.00

104.00
7×O

7×
16.00

+
112.00
261.98 g/mol Na2Cr2O7
15b)
Rewrite the molar mass of the substance above in conversion factor form:
261.98 g Na2Cr2O7 = 1 mol Na2Cr2O7
(Time Range: 1 min − all levels)
Page 1 of 13
EXAM 4 REVIEW
CHEM 110
Identify initial value!
16)
Calculate the number of molecules of NH3 in a 1.400 × 10−5 mole sample of NH3.
(Time Range: 3 to 6 min)
Solution/Explanation:
 You are converting between particles (molecules) and moles (mol). Therefore, you need to use Avogadro’s Number.

You need to devise a conversion factor: 1 mol NH3 = 6.022 × 1023 NH3 molec

Always start with your initial value: 1.400 × 10−5 NH3 molec

1 mol NH3 = 6.022 × 1023 NH3 molec
initial value
-5
1.400 × 10
(
1
NH3 mol
)
Avogadro’s Number Conversion
23
6.022 × 10
NH3 molec
(
) =
1 mol NH3
18
8.431 × 10
NH3 molec
Page 2 of 13
Identify initial value!
17)
A sample of CaCl2 contains 3.8 × 1025
of CaCl2 is 110.98 g/mol.
EXAM 4 REVIEW
CHEM 110
formula units of CaCl2. Calculate the mass (in grams) of CaCl2 in the sample. The molar mass
Identify final unit: mass (grams)
(Time Range: 3 to 5 min)
Solution/Explanation:
 You are converting between particles (molecules) and mass (grams). Therefore, you need to use Avogadro’s Number.

There is no direct pathway from particles to moles: mass  moles  particles

You must go through moles first. You need two conversion factors:

110.98 g CaCl2 = 1 mol CaCl2

6.022 × 1023 CaCl2 form = 1 mol CaCl2
initial value
(
3.8 × 10
25
1
Avogadro’s Number Conversion
CaCl2 form
)
1 mol CaCl2
(
)
23
6.022 × 10 CaCl2 form
=
63 mol CaCl2
molar mass
63 mol CaCl2
(
)
1
(
110.98 g CaCl2
)
1 mol CaCl2
=
7003 g CaCl2 
x7.0 × 103 g CaCl2x
Also acceptable:
initial value
(
3.8 × 10
25
1
CaCl2 form
)
Avogadro’s Number Conversion
molar mass
1 mol CaCl2
(
)
23
6.022 × 10
CaCl2 form
110.98 g CaCl2
(
)
1 mol CaCl2
=
7003 g CaCl2 
x7.0 × 103 g CaCl2x
Page 3 of 13
EXAM 4 REVIEW
CHEM 110
Reaction for Question 18
283.88 g/mol
P4O10(s)
63.01 g/mol
+
12 HNO3(aq)
98.00 g/mol
4 H3PO4(aq)

108.01 g/mol
+
6 N2O5(s)
Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5)
Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from
500.0 grams of nitric acid, HNO3(aq).
Identify initial value!
18)
(Time Range: 6 to 12 min)
Solution/Explanation:

Steps to Theoretical Yield Calculations

Step 1: Convert the initial value (the KNOWN) to moles

Step 2: Use mole ratios to convert moles KNOWN  moles UNK

Step 3: Convert moles UNK to mass UNK
molar mass
initial value
STEP 1:
*
(
500.0 g HNO3
)
1
(
1 mol HNO3
)
63.01 g HNO3
=
7.935 mol HNO3
Use molar mass of the balanced equation to convert mass of HNO3 to moles of HNO3.
mole ratio
STEP 2:
*
7.935 mol HNO3
(
)
1
6 mol N2 O5
(
)
12 mol HNO3
=
3.968 mol N2O5
Use the coefficients of the balanced equation to convert moles of HNO3 to moles of N2O5.
molar mass
STEP 3:
3.968 mol N2 O5
(
)
1
108.05 g N2 O5
(
)
1 mol N2 O5
=
x428.7 g N2O5 OR 4.287 × 102 g N2O5x
*
Convert moles of N2O5 to mass of N2O5 using molar mass.
*
You can always put answers to calculations in standard SCINOT. (Make sure the SCINOT is
correct.)
See next page to see “all-in-one-step” calculation:
Page 4 of 13
EXAM 4 REVIEW
CHEM 110
Reaction for Question 18
283.88 g/mol
P4O10(s)
63.01 g/mol
+
12 HNO3(aq)
98.00 g/mol
4 H3PO4(aq)

108.01 g/mol
+
6 N2O5(s)
Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5)
18)
Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from 500.0 grams of nitric acid, HNO3(aq).
Identify initial value!
(Time Range: 6 to 12 min)
Solution/Explanation:

Steps to Theoretical Yield Calculations

Step 1: Convert the initial value (the KNOWN) to moles

Step 2: Use mole ratios to convert moles KNOWN  moles UNK

Step 3: Convert moles UNK to mass UNK
INITIAL VALUE
STEP 1
STEP 2
STEP 3
initial value
molar mass
mole ratio
molar mass
1 mol HNO3
(
)
63.01 g HNO3
6 mol N2 O5
(
)
12 mol HNO3
108.05 g N2 O5
(
)
1 mol N2 O5
(
500.0 g HNO3
)
1
=
x428.7 g N2O5 OR 4.287 × 102 g N2O5x
* Again, I have no preference as to whether you break down this solution into three steps or choose the “all-in-one-step” method.
Page 5 of 13
19)
EXAM 4 REVIEW
CHEM 110
Assume that a scientist perform this experiment and obtains a yield of 411.7 grams of N2O5. Calculate the %-yield value for this
experiment. Use the value obtained in Question 18 for the theoretical yield.
Note: You must rewrite the %-yield equation first with variables only as your first step.
(Time Range: 2 to 4 min)
Solution
%-yield =
experimental yield
theoretical yield
× 100
=
=
experimental yield
theoretical yield
411.7 g N2 O5
428.7 g N2 O5
× 100
× 100
= 96.03 % yield N2 O5
Page 6 of 13
EXAM 4 REVIEW
CHEM 110
PART 3: NON-CALCULATOR


20)
Resources:

You will be given a periodic table (symbols only)

You will be given the memorized table of polyatomic ions chart

You will be given the Activity Series Table (Top Dog)

You will be given the Solubility Rules (all of them)

No calculators for this section
Timing: 10 min (mastery)  25 min (competence)
Indicate whether the following elements are solid (s), liquid (l), or gas (g).
20a)
Cu (#29)
(s)
20b)
Hg (#80)
(l)
(Time Range: 1 min maximum – all levels)
21)
List the charges of the following elements when they become ions:
21a)
nitrogen (#7)
−3 or 3−
21c)
potassium (#19)
+1 or 1+
21b)
sulfur (#16)
−2 or 2−
21d)
iodine (#53)
−1 or 1−
(Time Range: 1 min 30 s maximum – all levels)
IMPORTANT: For positive ions, you must write the “+” symbol!
Question 21a has been corrected. It should be −3 (not +3). Sorry.
Page 7 of 13
EXAM 4 REVIEW
CHEM 110
22)
Circle the metals that require a Roman Numeral in their names when they are present in
ionic compounds.
Na (#11)
Fe (#26)
Sn (#50)
Al (#13)
Zn (#30)
Ba (#56)
(Time Range: 2 min maximum – all levels)
Na
Group 1A (alkali metals) always have a charge of 1+. Therefore, Roman numerals are not
required when naming salts that contain Group 1A metals.
Fe
Transition metal. With three exceptions, transitions metals always require a Roman
numeral in their names to indicate their charge.
Sn
Post-Transition metal. All post-transition metals always require a Roman numeral in
their names to indicate their charge.
Al
Group 3A. This is a little tricky. Even though there are transition metals in Group 3A,
aluminum itself is not a transition metal. Therefore, it does not require a Roman numeral
to indicate its charge.
Zn
Zinc is one of the three exceptions of the transition metals. Zinc always has a charge
2+ in ionic compounds. Therefore, a Roman numeral is not required. Here are the three
exceptions again:



Ba
Ag (silver) is always 1+
Cd (cadmium) is always 2+ in ionic compounds
Zn (zinc) is always 2+ in ionic compounds
When present in ionic compounds,
these three metals do not require a
Roman numeral when naming them.
Group 2A (alkali earth metals) always have a charge of 2+. Therefore, Roman numerals
are not required when naming salts that contain Group 2A metals.
Page 8 of 13
EXAM 4 REVIEW
CHEM 110
23)
Circle the compounds that are soluble in water.
potassium sulfate
V2O5
ammonium phosphate
CuCl2
chromium(III)
hydroxide
Au2(Cr2O7)3
(Time Range: 3 min maximum – all levels)
potassium sulfate
K2SO4
potassium
V2O5
vanadium(V) oxide
vanadium(V)
K+

Rule 1 = soluble

Rule 6 = soluble
V5+

Rule 7 = insoluble
oxide
O 2−

Rule 5 = insoluble
ammonium phosphate
ammonium
NH4 +

Rule 1 = soluble
(NH4)3PO4
phosphate
PO4 3−

Rule 5 = insoluble
CuCl2
copper(II) chloride
copper(II)
Cu2+

Rule 7 = insoluble
chloride
Cl −

Rule 4 = soluble
chromium(III)
Cu2+

Rule 7 = insoluble
Cr(OH)3
hydroxide
OH−

Rule 5 = insoluble
Au2(Cr2O7)3
gold(III)
Au3+
Rule 7 = insoluble
dichromate
Cr2O7 2−
Rule 7 = insoluble
chromium(III) hydroxide
Gold(III) dichromate
sulfate
SO4
2−
Rule 1 beats Rule 6
Rule 5 beats Rule 7
Rule 1 beats Rule 5
Rule 4 beats Rule 7
Rule 5 beats Rule 7
Rule 7 (tie)
Page 9 of 13
EXAM 4 REVIEW
CHEM 110
24)
Balance the following two equations:
2
24a)
C4H10(l)
+
9
O2(g)
∆
→
8
CO(g)
+
10
H2O(g)
Note: For Question 24a, I purposely did not tell you how much oxygen was present. It does not matter, since the products are
already written. Sorry about the wrong answer.
3
24b)
Hint:
Time Limit:
25)
Ca(OH)2(s)
+
2
H3PO4(aq)

1
Ca3(PO4)2(s)
+
6
HOH(g)
Rewrite water as _____ to help you balance.
4 minutes maximum – all levels.
Write the skeleton equation: (Don’t forget states: s, l, g, aq)
phosphorus(s) + chlorine(g)  phosphorus pentachloride(s)
P(s) + Cl2(g)  PCl5(s)
Time Limit:
26)
2 minutes to 4 minutes
Write the word equation: (Don’t forget states: s, l, g, aq)
Cl2O7(s) + H2O(l)  HClO4(aq)
dichlorine heptoxide(s) + water(l)  perchloric acid(aq)
Time Limit:
2 minutes to 4 minutes
Page 10 of 13
EXAM 4 REVIEW
CHEM 110
27)
Predict the products (chemical formulas):

Do not balance equations.

Include states: (s), (l), (g), or (aq)

Hint: Name the products first. Write the names in the margins. Then write the chemical formulas from the names.
FeBr3(aq) + Cl2(g)  FeCl3(aq) + Br2(l)
27a)
Time Limit: 2 minutes maximum – all levels
Solution

The reactants are: ionic compound + an element. Therefore, this is a Single Replacement (SR)

Before anything else, name the reactants: iron(III) bromide + chlorine (g) 

The element (chlorine gas) is a nonmetal. Therefore, it will try to replace bromide.

Check the Activity Series Table (Appendix 8): chlorine beats bromine. Therefore, chlorine and bromide switch places.

Before anything else: name the products:
iron(III) bromide
+
FeBr3(aq)
chlorine

iron(III) chloride
+
bromine
Cl2(g)

FeCl3(aq)
+
Br2(l)
Note that the suffixes –ine and –ide stay put.

Write the chemical formulas from the names.

iron(III) chloride is a salt. Therefore, you must determine if water is hanging around, and if iron(III) chloride is soluble.


The “aq” in the reactants tells you that water is present. Therefore, you must determine the solubility of iron(III) chloride.

Solubility Table, Rule 4, chlorides are soluble. Therefore, iron(III) chloride is aqueous.
Don’t forget states


iron(III) chloride

The “aq” in the reactants tells you that water is present. Therefore, you must determine the solubility of iron(III) chloride.

Solubility Table, Rule 4, chlorides are soluble. Therefore, iron(III) chloride is aqueous.
bromine

The states of all elements should be memorized. Elemental bromine, Br2, is a liquid.
Page 11 of 13
EXAM 4 REVIEW
CHEM 110
XS
27b)
C5H8(l)
+
O2(g)
∆
→
CO2(g) + H2O(g)
Time Limit: 1 minute maximum – all levels
Solution

C5H8 is a hydrocarbon (contains carbon and hydrogen).

The other reactant is oxygen, and the reaction is being heated.

Therefore, this is a HYDROCARBON COMBUSTION reaction.

Since there is excess oxygen, the products are carbon dioxide gas and water vapor (memorized). There is no way to “figure this out.” You
either have learned/memorized this, or not.
K(s) + Br2(l)  KBr(s)
27c)
Time Limit: 1 minute maximum – all levels
Solution

This is a simple COMBINATION reaction. Metal + nonmetal  salt

Name the salt: potassium bromide (common sense)

Write the chemical formula from the name.

States


If a product is a salt (ionic compound), the salt is a solid unless water is hanging around.

There are no “aq” in the reactants, water is NOT a reactant, and water is NOT present over the reaction arrow →

Water is not present, so you do not need to check for solubility. This product is a solid.
H2 O
Do not worry that the equation is not balanced. I asked for the unbalanced equation.
Page 12 of 13
EXAM 4 REVIEW
CHEM 110
28)
The following reaction is NR. In three or less sentences, explain why no reaction occurs.
Time Limit: 3 minutes maximum – all levels
(NH4)3PO4(aq) + K2SO4(aq)  (NH4)2SO4(aq) + K3PO4(aq)
This is a DOUBLE REPLACEMENT (DR) reaction. In DR reactions, if all substances (reactants and products) are soluble
(aqueous), then there is no reaction. All reactants and products in the reaction above are aqueous, so NR.
Page 13 of 13