Practice Homework 24: Solubility Practice
Definitions/Helpful Info
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You will have access to the full rules for determining solubility.
Do not think that – just because you are getting the solubility rules – you don’t have to practice. The test
is long, and you do not need to waste time flipping through the rules for the first time. You will run out of
time.
As a result, you might want to memorize the first two rules.
The Solubility Rules are located on the website as Appendix 09 - Solubility Rules - NVC
Fall 2015 students: I apologize for not having 10 practice sets for this problem. My mind is swiss cheese
right now.
Instructions
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Circle the substances that are soluble in water.
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Underneath each substance, indicate the rule number that tells you that the substance is soluble.
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You will not have to indicate the rule number (reasoning) on Exam 4.
You will need your own paper for this set of problems.
Problem Set 1
PbSO4
calcium acetate
Na2CO3
chromium(III) nitrate
LiOH
iron(II) dichromate
BaCl2
zinc sulfide
KI
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Problem Set 2
ammonium iodate
TiO2
calcium nitride
AgBr
potassium persulfate
Cu(OH)2
lithium arsenide
ZnSO4
lead(II) chromate
BaI2
nickel(III) acetate
Li3P
magnesium nitride
K2SO3
aluminum fluoride
Ca(NO2)2
sodium hydrogen
carbonate
SnCO3
Problem Set 3
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Answers
Problem Set 1 – Answer Keys
PbSO4
calcium acetate
Na2CO3
Rule 3
Rule 2
Rule 1
chromium(III) nitrate
LiOH
iron(II) dichromate
Rule 2
Rule 1
Rule 7
BaCl2
zinc sulfide
KI
Rule 4
Rule 5 (S2−)
Rule 1
ammonium iodate
TiO2
calcium nitride
Rule 1
Rule 5
Rule 7
AgBr
potassium persulfate
Cu(OH)2
Rule 3
Rule 1
Rule 5
lithium arsenide
ZnSO4
lead(II) chromate
Rule 1
Rule 6
Rule 7
Problem Set 2
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Problem Set 3
BaI2
nickel(III) acetate
Li3P
Rule 5
Rule 2
Rule 1
magnesium nitride
K2SO3
aluminum fluoride
Rule 7
Rule 1
Rule 7
Ca(NO2)2
sodium hydrogen
carbonate
SnCO3
Rule 7
Rule 1
Rule 5
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166.00 g/mol
7)
6 KI(aq)
18.02 g/mol
+
6 H2O(l)
158.03 g/mol
+
253.80 g/mol

2 KMnO4(aq)
3 I2(s)
86.97 g/mol
+
MnO2(s)
56.11 g/mol
+
8 KOH(aq)
7a)
Calculate the mass of manganese(IV) oxide that can be synthesized from 15.00 grams of potassium iodide.
7b)
Calculate the percent yield of this experiment if a mass of 1.982 grams of manganese(IV) oxide is produced.
394.72 g/mol
8)
2 NI3(s)
28.02 g/mol

1 N2(g)
253.80 g/mol
+
3 I2(s)
8a)
Calculate the theoretical yield of iodine, assuming an initial mass of 80.00 g of nitrogen triiodide.
8b)
Calculate the percent yield, if a mass of 70.00 grams of iodine is obtained in an experiment.
Question 9
277.13 g/mol
Reaction 9a)
2 C7H5N3O6(s)
28.02 g/mol

277.13 g/mol
Reaction 9b)
2 C7H5N3O6(s)
3 N2(g)
18.02 g/mol
+
28.02 g/mol

3 N2(g)
5 H2O(g)
28.01 g/mol
+
2.02 g/mol
+
H2(g)
7 CO(g)
12.01 g/mol
+
28.01 g/mol
+
12 CO(g)
7 C(s)
12.01 g/mol
+
2 C(s)
The compound C7H5N2O6 is trinitrotoluene (TNT). TNT can explosively decompose from either of the reactions above. Under controlled
conditions, 5.00 grams actually yielded 2.97 grams of carbon monoxide gas.
Question 9a: Calculate the theoretical yield of carbon monoxide gas by through both reactions.
Question 9b: Determine whether the reaction occurred by Reaction 9a or Reaction 9b.
Question 9c: Determine the theoretical yield of this reaction.
Answer Key will be up by tomorrow afternoon.
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Answers
1a)
32.95 g AlCl3
1b)
76.58%-yield AlCl3
2a)
4.30 g HClO4
2b)
94.9%-yield HClO4
3a)
60.58 g CO2
3b)
91.96%-yield CO2
4a)
32.11 g F2
4b)
75.43%-yield F2
5a)
294.23 g V2O3
5b)
62.265%-yield V2O3
6a)
5.972 g PH3
6b)
103%-yield PH3
6c)
The extra mass is probably due to unreacted water still in the product.
7a)
77.16 g MnO2
7b)
90.72%-yield MnO2
8a)
2.620 g I2
8b)
75.66%-yield I2
9a)
Reaction 9a produces 2.16 g CO. Reaction 9b produces 3.70 g CO
9b)
As 2.97 g CO was produced experimentally, the reaction most likely proceeded through
Reaction 9b.
9c)
As the reaction proceeded through Reaction 9b, the percent yield of CO would be 85.5%
Page 6 of 6