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PRACTICE EXAM III (Ch. 8-10) 5th and 6th edition of Corwin’s
The contents of these chapters are more calculation-oriented and are the beginning of
learning of the chemical language. More study is required to comprehend the
concepts. Probably the most difficult test among the four.
Multiple Choices: Choose the best (one) answer. Show in bold.
Questions break-down: Chapter 8: Q1-8; Chapter 9: Q9-16: Chapter 10: 17-25.
Multiple Choices: Choose the best answer. The correct answer is shown in bold
character.
1. Which of the following is evidence for a chemical reaction?
(A) A gas is detected.
(B) A precipitate or solid is formed.
(C) A color change is observed.
(D) An energy change is noted.
(E) All of the above.
Hint: See both 5th ed. and 6th ed. Section 8.1
2. Which of the following statement best describes the reaction below?
Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)
(A) Lead nitrate reacts with potassium chromate to give lead chromate and
potassium nitrate.
(B) Lead (II) nitrate reacts with potassium chromate to give lead (II) chromate
and potassium nitrate.
(C) Lead nitrate solution reacts with potassium chromate solution to give solid
lead chromate and potassium nitrate solution.
(D) Lead (II) nitrate aqueous solution reacts with potassium chromate aqueous
solution to give solid lead (II) chromate and potassium nitrate aqueous
solution.
Hint: See both 5th ed. and 6th ed. Section 8.2. Also transition metal, lead, requires
specify its charge as described in Sections 7.4 and 7.6.
3. If the equation C4H10 + O2  CO2 + H2O is balanced, which of the following
quantities is correct? Note: Coefficients must be integers.
(A) 2 C4H10, 13 O2, 8 CO2, and 10 H2O
(B) 13 C4H10, 2 O2, 10 CO2, and 8 H2O
(C) C4H10, 6.5 O2, 4 CO2, and 5 H2O
(D) C4H10, 2 O2, 5 CO2, and H2O
Hint: See both 5th ed. and 6th ed. Section 8.3. However, you must read Section
8.2 first to be able to translate a chemical reaction into a chemical equation
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when the question is given as a description. For instance, the above question
can be given as
“Balance the equation when butane gas reacts with oxygen gas (combustion or
burning)” or “write a balanced chemical equation when butane reacts with
oxygen”.
Balancing equation is a very, very important question. To balance a chemical
equation, you must make sure the number of atoms of each kind at both sides
of the arrow is identical. Also start examining the most bulky species, that is, the
one with the most different kinds of atoms and number of atoms.
In this question, C4H10 is the most bulky one. So we use it as reference by putting
1 in front of it to remind us we have done examining C4H10. Now the equation is
updated to be 1 C4H10 + __ O2  __ CO2 + __ H2O
Since C4H10 contains 4 carbon atoms, so we need four carbon atoms at the right
side, which leads us to put 4 (called coefficient) in front of the CO2.
Now the equation is updated to be 1 C4H10 + __ O2  4 CO2 + __ H2O
As there are 10 hydrogen atoms in C4H10, thus we need to balance the hydrogen
atoms, which lead us to put 5 in front of the H2O.
Now the equation is updated to be 1 C4H10 + __ O2  4 CO2 + 5 H2O
Now we need to balance the oxygen atoms. Since there are 4x2+5x1 = 13 oxygen
atoms at the right side, and thus the left side must have the same number. That
is __x2 = 13. So __ = 6.5
According to the rule, all the coefficients must be whole numbers and you
cannot round them, so we must multiply 6.5 by a whole number, which must be
2 or greater, until it reaches an integer. We find that 2 will do the job. So each
coefficient must be multiplied by 2 as well. That is,
2 x (1 C4H10 + 6.5 O2  4 CO2 + 5 H2O) which leads to 2 C4H10 + 13 O2  8 CO2 +
10 H2O
Thus, we have balanced this equation.
4. Which of the following statement is wrong?
(A) AgNO3
3 (aq) is a precipitation reaction.
(B) NH4
3 (g) + HCl (g) is a decomposition reaction.
(C) 2C4H10 + 13O2  8CO2 + 10H2O is a combustion reaction.
(D) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) is a neutralization reaction.
E) Zn(s) + 2 AgNO3(aq) 2 Ag(s) + Zn(NO3)2(aq) is a double-replacement
reaction
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Hint: See both 5th ed. and 6th ed. Section 8.4 for summary and Sections 8.5 and
8.6 for examples. Please note that Choice (A) is also called a double-replacement
reaction or a metathesis reaction.
5. What is the product predicted from the following combination reaction?
Al(s) + O2 (g) 
(A) AlO
(B) AlO2
(C) Al2O3
(D) Al3O2
(E) Al2O
Hint: See both 5th ed. and 6th ed. Section 8.5. Please note that reaction with
oxygen gas is also called combustion reaction or oxidation-reduction (i.e. redox)
reaction.
What are the products from the following decomposition reaction?
CaCO3 (s) 
(A) CaO (s) and CO2 (g)
(B) Ca(s) and CO3(g)
(C) CaO2 (g) and CO(g)
(D) CaCO and O2 (g)
(E) 2 → 1
Hint: See both 5th ed. and 6th ed. Section 8.6. Please note that the decomposition
reaction is the reverse reaction of combination.
6. Which of the following metals reacts with aqueous FeSO4?
Partial Activity Series: Mg > Al > Zn > Fe > (H)
(A) Al
(B) Mg
(C) Zn
(D) All of the above.
(E) None of the above.
Hint: See both 5th ed. and 6th ed. Section 8.7. Information (activity series for
metals) at the margin of p. 214, 6th ed. is very useful. The reaction direction in p.
214 is always from left to right.
7. What are the products from the following single-replacement reaction?
Na(s) + H2O(l) 
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(A) Na2O and H2
(B) Na2O and H2O
(C) NaOH and H2
(D) NaOH and H2O
(E) No reaction.
Hint: See both 5th ed. and 6th ed. Section 8.8. Watch out the charges of ions
before combining them into an electrically neutral compound or element.
What are the products from the following double-replacement reaction?
Pb(NO3)2 + K2CrO4 
(A) Pb(NO3)2 and K2CrO4
(B) K2CrO4 and PbCrO4
(C) PbCrO4 and KNO3
(D) Pb(NO3)2 and KNO3
(E) All of the above.
Hint: See both 5th ed. and 6th ed. Section 8.10. Watch out the charges of ions
before combining them into an electrically neutral compound or element.
8. Which of the following solid compounds is insoluble in water?
(A) Ba(OH)2
(B) AgNO3
(C) PbCl2
(D) (NH4)2CO3
(E) Ca(NO3)2
Hint: See both 5th ed. and 6th ed. Section 8.9. Know how to use Table 8.2, p. 219
sixth ed. This is a very important question. Refer to the Solubility Rules in
Appendix E in the 5th ed. and Appendix D, p. 600, in the 6th ed. of the textbook.
Note that Ca(OH)2 is soluble in water (textbook is correct) while CaSO4 is
insoluble in water (Please correct your textbook Table 8.2 as your textbook said
it’s soluble.
9. Which of the following statement is correct?
(A) 1 mole of Ca atoms contains 6.02 x 1023 Ca atoms.
(B) 3.01 x 1023 N2 molecules is equivalent to 0.5 mole of N2 molecules.
(C) The Avogadro’s number is 6.02 x 1023.
(D) There are 6.022 x 1023 NaCl formula units in 1 mole of NaCl.
(E) All of the above.
Hint: See both 5th ed. and 6th ed. Section 9.1 for Avogadro’s number and Section
9.2 for mole calculation.
10. What is the approximate molar mass of calcium nitrate, Ca(NO3)2 ?
(A) 164.1 g/mol
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(B) 106.4 g/mol
(C) 52 g/mol
(D) 98.5 g/mol
Hint: See both 5th ed. and 6th ed. Section 9.3. In the formula, Ca(NO3)2, it
indicates there are 1 Ca, 1x2 = 2 N and 3x2 = 6 O. Go to the periodic table and
locate the atomic mass for each atom. Thus the formula mass of Ca(NO3)2 = 1x40
+ 2x14 + 6x16 = 164. Note that if the unit given as amu (referring to atom mass
unit it refers to one Ca(NO3)2; if the unit given as grams or g/mol it refers to
contains 6.02 x 1023 Ca(NO3)2 which is commonly called molar mass as 1 mole of
substance contains Avogadro number of particles (atoms or molecules) or units.
11. A gas has a density of 3.84 g/L at a pressure of 3.50 atm and a temperature of
37oC. What is the molar mass (g/mol) of this gas?
(A) 27.89
(B) 0.102
(C) 497.28
(D) 58.73
Hint: See both 5th ed. and 6th ed. Sections 9.5 and 9.6. Be very careful that the
method of Example Exercise 9.9 works only for 1 L gas at STP (0 oC and 1 atm)
condition. This question the gas is at 3.5 atm and 37oC and thus it does not
apply.
The easiest way to solve this question not under the standard condition is to
apply the derived ideal gas equation: PM = dRT.
Here, P is pressure in atm, M is molar mass in gram/mol, d is density in
gram/liter, R is the ideal gas constant, 0.082 atm.l/mol.K, T is temperature in
Kelvin. Recall K = oC + 273.15.
This equation, PM = dRT, is derived from the ideal gas equation PV = nRT where
P is pressure in atm, V is volume in leter, n is mole, R is the ideal gas constant,
0.082 atm.l/mol.K, and T is temperature in Kelvin.
Since n = mole = mass in gram /molar mass, we can re-write PV = nRT to
PV= (m/M)RT where m is the mass in gram and M is the molar mass in
gram/mole.
By switching the positions of V and M, we can re-write the equation to
PM = (m/V)RT = dRT where d is the density in gram/liter.
So 3.50 x M = 3.84 x 0.082 x (37 + 273.15) = 97.66  M = 97.66/3.50 = 27.90
g/mole
12. What is the mass of 0.500 Liter of oxygen gas, O2, at STP?
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(A) 0.286 g
(B) 0.714 g
(C) 3.50 g
(D) 6.40 g
(E) 112 g
Hint: See both 5th ed. and 6th ed. Sections 9.5 and 9.6. The method shown in the
textbook Example Exercise 9.10 is (0.500/22.4) x 32 = 0.714 g as at STP one mole
of oxygen gas weighs 32 grams and occupies 22.4 liters.
STP refers to standard temperature and pressure, that is, 0oC and 1 atm. The
molar mass of O2 is 32.0 g/mol. There are two other ways to solve this question:
one is using PM = dRT and another is applying PV=nRT and n=mass/molar
mass.
Using PM=dRT method:
1 x 32 = d x 0.082 x (0+273.15)  d = 32/ 22.3983 = 1.43 g/L. So the mass of
oxygen gas = 1.43 x 0.500 = 0.714 g.
Applying PV=nRT and n=mass/molar mass method:
1 x 0.5 = n x 0.082 x 273.15 n = 0.5/22.3983 = 0.00223 mole
Mass = mole x molar mass = 0.00223 x 32 = 0.714 gram
13. What is the percentage of oxygen in carbon dioxide, CO2?
(A) 12.37%
(B) 27.27%
(C) 36.36%
(D) 57.14%
(E) 72.73%
Hint: See both 5th ed. and 6th ed. Section 9.7 Example Exercise 9.12.
O% = 2 x {16.00/(1x12+2x16)} x 100% = {32/44}x100% = 72.73%.
% composition of an atom = n x (molar mass of an atom/molar mass of
compound) x 100%
14. If 0.250 mol V reacts with 0.375 mol O, what is the empirical formula of
vanadium oxide?
(A) VO
(B) V2O3
(C) V2O5
(D) V3O2
(E) V5O2
Hint: See both 5th ed. and 6th ed. Section 9.8. Empirical formula must be
obtained from experiment and it is simplest whole number mole ratio of atoms.
V: O = 0.250 : 0.375 = (0.250/0.250) : (0.375/0.250) = 1 : 1.5 = 1x2 : 1.5x2 = 2 : 3
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Do not round the numbers for no valid reasons. Note that 0.5 = 1/2; 0.33 or 0.34
= 1/3; 0.25 = ¼.
15. What is the empirical formula for methyl benzoate, a compound used in the
manufacture of perfumes, contains 70.57% carbon, 5.94% hydrogen, and 23.49%
oxygen? Note: subscripts must be integers.
A) C4H4O
B) C2H2O0.5
C) C8H8O2
D) CHO
Hint: Section 9.8 Example Exercise 9.14 (sixth edition). This is a very, very
important question. The empirical formula is the simplest integral ratio of moles
among each atom. Here, there are three different kinds of atoms, C, H and O.
Thus mole of C = 70.57/12 = 5.88; mole of H = 5.94/1 = 5.94; mole of O =
23.49/16 = 1.47.
Note that as long as one of the moles is not an integer, we have to divide the
smallest value among them: here the smallest value is 1.47. So C : H : O =
5.88/1.47 : 5.94/1.47 : 1.47/1.47 = 4 : 4.04 : 1. Since 4.04 is very close to 4.00 and
thus we can round it to 4.00. So C : H : O = 4 : 4 : 1, which indicates that the
empirical formula contains four C, four H and one O. Thus, the empirical formula
is written as C4H4O as 1 is usually not written in the formula.
*******There is a type of question that requires two steps to solve it. See the
following vitamin C question.*******
The composition of vitamin C or ascorbic acid is made by 40.92% C, 4.58% H and
54.50% O. What is the empirical formula for vitamin C?
C : H : O = 40.92/12.01 : 4.58/1.01 : 54.50/16.00 = 3.407 : 4.54 : 3.406 = 1: 1.33 :
1 = 3 : 4 : 3.
As long as one of the values is not an integer, we need to modify them by
multiplying a whole number 2 or greater than 3 until all the numbers turn to be
whole numbers. Since multiply by two will not make 2.66 as an whole number,
and thus we try to multiply by three. This time it works.
Please note that we can NOT round 3.407 or 3.406 to 3 or to 4 as it will cause
too much error. Same as for 4.54 can NOT be rounded to 4 or 5.
16. The empirical formula of ethylene glycol is CH3O. The weight of 1 mole of
ethylene glycol is 62.1 g. What is its molecular formula? Note: subscripts must
be integers.
A) C2H6O2
B) CHO
C) CH3O
D) C0.5H0.5O0.5
Hint: See both 5th ed. and 6th ed. Section 9.9 Example Exercise 9.15 (sixth
edition). Remember the definition that the molecular formula is an integral
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multiple of empirical formula. That is, the molar mass (i.e. molecular weight) =
empirical formula weight x integer. So the empirical formula weight of CH3O =
12x1+1x3+16x1=31. So the integer = 62.1/31 = 2. Thus there are two empirical
formulas in a molecular formula. Therefore, the molecular formula is C2H6O2.
17. How many moles of oxygen gas, O2, react with 2 moles of nitrogen monoxide gas,
NO, according to the following equation?
__ NO(g) + __ O2(g)  __ NO2(g)
(A) 1 mol.
(B) 2 mol.
(C) 3 mol.
(D) 4 mol.
(E) None of the above.
Hint: See both 5th ed. and 6th ed. Section 10.1 in-text example. However, to be
more accurate, this question involves concepts from Sections 10.1 to 10.3:
balancing equation and stoichiometry, which deals with mole–mole relationship.
As the coefficients (the numbers proceed chemical formulas; shown in blanks to
be balanced in the above question) represent the mole–mole relationship.
18. How many moles of water react with 0.500 mol of lithium metal according to the
following reaction?
__ Li(s) + __ H2O  __ LiOH(aq) + __ H2(g)
(A) 0.125 mol.
(B) 0.250 mol
(C) 0.500 mol.
(D) 1.00 mol
(E) 2.00 mol.
Hint: See both both 5th ed. and 6th ed. Section 10.2. First, balance the equation.
Then apply mole-mole relationship according to the balance equation.
Balanced equation:
_2_ Li(s) + _2_ H2O  _2_ LiOH(aq) + _1_ H2(g)
Since the ratio for Li and H2O is 2 : 2 = 1 : 1 ratio, so the mole of water required
will be the same mole number of Li consumed.
18. How many moles of water must react in order to produce 0.500 mol of
potassium hydroxide?
__ K(s) + __ H2O(l)  __ KOH(aq) + __ H2(g)
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(A) 0.125 mol.
(B) 0.250 mol
(C) 0.500 mol.
(D) 1.00 mol
(E) 2.00 mol.
Hint: See both 5th ed. and 6th ed. Section 10.2. Same concept as Q 17. First,
balance the equation. Then apply mole-mole relationship according to the
balance equation.
_2_ K(s) + _2_ H2O(l)  _2_ KOH(aq) + _1_ H2(g)
Since the ratio for KOH and H2O is 2 : 2 = 1 : 1 ratio, so the mole of water
consumed will be the same mole number of KOH produced.
20. How many grams of aluminum metal must react to give 500.0 g of iron?
__ FeO(l) + __ Al(l)  __ Fe(l) + __ Al2O3
(A) 80.6 g
(B) 161 g
(C) 242 g
(D) 362 g
(E) 483 g
Hint: See both 5th ed. and 6th ed. Section 10.4: Mass-Mass problems. First,
balance the equation. Then apply mole-mole relationship according to the
balance equation. Then convert mole to mass.
_3_ FeO(l) + _2_ Al(l)  _3_ Fe(l) + _1_ Al2O3
Mole of Fe = (500.0/55.845) = 8.953 mol
{2 Al/ 3 Fe} = {y mol Al/8.953 mol Fe} => y = 2x8.953/3 = 5.969 mol Al
Grams Al = 5.969 mol Al x 26.982 g/mol = 161.05 g Al
Remember that there is no direct mass relationship in chemical reactions. See
p. 284 (sixth edition) Concept Map—Summary of Stoichiometry or the Road
Map given by the instructor.
21. How many grams of mercuric oxide (216.59 g/mol) must decompose to release
0.375 L of oxygen gas at STP?
__HgO(s)  __ Hg(l) + __ O2(g)
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(A) 0.0752 g
(B) 1.82 g
(C) 3.63 g
(D) 7.25 g
(E) 14.5 g
Hint:See both 5th ed. and 6th ed. Section 10.5: Mass-Volume problems. First,
balance the equation. Then apply PV = nRT to calculate mole of oxygen. STP
means 1 atm and 0oC. Then apply mole relationship to calculate mol of HgO.
Then cover mol of HgO to mass of HgO.
_2_HgO(s)  _2_ Hg(l) + _1_ O2(g)
1 x 0.375 = n x 0.082 x (0 + 273.15)  n = 0.375/22.3983 = 0.01674 mol O2
2 HgO/ 1 O2 = y mol HgO / 0.01674 mol O2  y = 2 x 0.01674 = 0.03348 mol HgO
Grams of HgO = 0.03348 mol HgO x 216.59 gram/mol = 7.2514 grams HgO
22. Give the following equation, 3 BaCl2 + 2 Na3PO4  Ba3(PO4)2 + 6 NaCl, how many
moles of Na3PO4 will react with 0.45 mole of BaCl2?
(A) 0.3
(B) 13.3
(C) 3.5
(D) 4.6
(E) 2.4
th
th
Hint: See 5 ed. Section 10.5 and 6 ed. Section 10.2: Mole-Mole problems.
From the equation, the involving species, 3 BaCl2 and 2 Na3PO4 with coefficients
3 and 2 respectively, tell us that for 3 moles of BaCl2 it requires 2 moles of
Na3PO4. According to this proportion or ratio, 0.45 mole of BaCl2 requires 0.45 x
2/3 = 0.3 mole of Na3PO4.
23. If 1.00 mol of chromium reacts with 1.00 mol of oxygen gas according to the
following reaction, how many moles of chromium (III) oxide are produced?
__Cr(s) + __ O2 (g)  __ Cr2O3(s)
(A) 0.500 mol
(B) 0.667 mol
(C) 1.00 mol
(D) 1.50 mol
(E) 2.00 mol
Hint: See both 5th ed. and 6th ed. Sections 10.7 and 10.8 Example Exercise 10.9:
The limiting reactant concept. It’s very, very important. First, balance the
equation. Second apply the stoichiometric (that is mole-mole relationship
according to the balanced equation).
_4_Cr(s) + _3_ O2 (g)  _2_ Cr2O3(s)
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If oxygen is the limiting reactant, then
{3 O2/ 2 Cr2O3} = {1 mol O2 / y mol Cr2O3}  y = 0.667 mol Cr2O3
produced.
If Cr is the limiting reactant, then
{4 Cr/ 2 Cr2O3} = {1 mol Cr / z mol Cr2O3}  z = 0.500 mol Cr2O3 produced.
Since 0.500 mol < 0.667 mol, thus there will be 0.500 mol Cr2O3 actually
produced and Cr is the limiting reactant and oxygen is the excess reactant.
24. If 82.4 g of aluminum metal (M=26.98 g/mol) reacts with 117.65 g of oxygen gas
(M=32.00 g/mol) according to the following equation, how many grams of aluminum
oxide (M=101.96 g/mol) are produced?
.
4 Al(s) + 3 O2(g)  2 Al2O3(s)
(A) 155.7 g
(B) 250.0 g
(C) 308.0 g
(D) 350.4 g
Hint: See both 5th ed. and 6th ed. Section 10.8 Example Exercise 10.10: Limiting
reactant concept.
Mole of Al = {82.4/26.98} = 3.054 mol Mole of oxygen gas = {117.65/32.00} = 3.677
mol
*If Al is the limiting reactant, then
4 Al /2 Al2O3 = 3.054 mol Al / y mole Al2O3  y = (2x3.054)/4 = 1.527 mol Al2O3
*If oxygen gas is the limiting reactant, then
3 O2 /2 Al2O3 = 3.677 mol O2 / z mole Al2O3  y = (2x3.677)/3 = 2.451 mol Al2O3
Since 1.527 mol Al2O3< 2.451 mol Al2O3, there will be actually 1.527 mol Al2O3
produced and Al is the true limiting reactant.
Grams of Al2O3 = 1.527 mol x 101.96 g/mol = 155.7 g. This is called the theoretical
yield in Section 10.9 because it is calculated from a balanced chemical equation.
The actual yield must be physically weighed by the scale or balance.
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25. Starting with 0.657 g of lead (II) nitrate, a student collects 0.905 g of precipitate.
If the calculated mass of precipitate is 0.914 g (i.e. theoretical yield), what is the
percent yield?
(A) 71.9%
(B) 72.6%
(C) 99.0%
(D) 101%
(E) 138%
Hint: See both 5th ed. and 6th ed. Section 10.9. Actual yield is the one that must
be physically weighed.
Percent yield = (0.905 g/ 0.914 g) x 100% = 99.02%
Note that there is no unit in the percent yield.
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