Page 1
Exam 2 practice. NOTE: Most (but not all) of these questions were actual questions on prior exams in CHM121.
However, I have patched them together from various exams, so the relative points won’t add up to 100 or necessarily
reflect the proportion of points on each topic for YOUR Exam 2. Make sure that you use the problem sets and
worksheets as your ultimate guide to what is "fair game". It looks to me like on this exam titration type calculations and
the concept of molecular formula as it relates to empirical formula were deemphasized compared to the problem set
problems assigned in this semester. –Prof. Mines
1. Circle the correct answer or provide an answer (2 points each)
(a) Which of the following is likely to be the most soluble in water?
ii) AgCl
i) CaCO3
iii) Fe2O3
iv) (NH4)2MoO4
Rule #2: Ammonium salts are soluble. Rule #3: AgCl is an exception (insoluble chloride). Rule #6: carbonates are insoluble unless paired with cations from Rule #2. No rule addresses oxide or iron(III), so you can’t assume soluble.
(b) In a 0.40 M aqueous solution of Li3PO4 the concentrations of Li+ and PO43- are:
3
4
4

O
P
M
0
4
.
0
=
3
4
 0.40 mol Li3PO4
x
L

Work: 

1.2 mol Li+
 1.2 M Li+ 
L

O
P
l L
o
m
0
4
.
0
3‐
[PO43-] = 0.40 M PO4

3 O
4
P
i3
L
l
o
m
1
 0.40 mol Li3PO4
x
L

Work: 
O
P
4l
O o
P m
i3 1
L
l
o
m
1
+
+
i
L
l
o
m
3
[Li+] = 1.2 M Li
(c) If 4.0 moles of Ca3(PO4)2 and 4.0 moles of SiO2 are put in a container, what will be the limiting reactant if reaction occurs as completely as possible according to the equation below?
2 Ca3(PO4)2 + 6 SiO2  6 CaSiO3 + P4O10
i) Ca3(PO4)2
ii) SiO2
iii) CaSiO3
iv) P4O10
v) none of the above
Qualitative reasoning: One needs 6 moles of SiO2 for every 2 moles of Ca3(PO4)2 for reaction to occur. That means you need more SiO2 (in moles). So if you start with equal moles, you will not have enough SiO2. Quantitative reasoning: Ratio PRESENT = 4 mol SiO2
4 mol Ca3 (PO4 )2
(=1) < Ratio in which they REACT =
6 mol SiO2
2 mol Ca3 (PO4 )2
(=3) Thus, there is not enough of the “numerator” present (to react with all of the “denominator”)  numerator (SiO2) is limiting. You could also do the “assumption” method or “max amount of product that could be made” approaches here as well. (d) When any number of moles of BrCl3 reacts according to the equation below:
2 BrCl3  3 Cl2 + Br2
i) The number of moles of Br2 formed is less than the moles of BrCl3 reacted.
ii) The number of moles of Cl2 formed is two-thirds of the moles of BrCl3 reacted.
iii) Three moles of Cl2 are formed.
iv) All of the above.
v) (i) and (ii) only.



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Looking at the coefficient ratio from the equation, 1 mol Br2 is formed for every 2 mol BrCl3 that reacts. That means fewer moles of Br2 are formed (“half” as many) than the # of moles of BrCl3 that react. RE: ii), the number of moles of Cl2 formed is actually three‐halves the # mol of BrCl3 reacted. RE: iii), That “3” coefficient represents part of a ratio only. It is not an “actual” amount. Clearly if you start with a greater or lesser amount of reactants, you will get a greater or lesser amount of products to form. It cannot be “fixed” by the symbols on a piece of paper! (e) The diagram below represents a closed container before any reaction occurs.
Legend
2 SO2(g) + O2(g)  2 SO3(g)
S=
O=
To see what (if anything) is left over, use the equation like a “recipe”: For every 2 molecules of SO2 that react, 1 O2 reacts with it. Using the approach noted in Problem 1 on PS5 (see key), I will circle each “equation unit” of reaction that occurs. Also make sure to represent the product properly: product is SO3, NOT “2 SO3” or S2O6.
Circle the diagram that shows what this actual container would look like after reaction occurs as
completely as possible according to the equation shown above:
(i)
(ii)
(iii)
(iv)
(v)
(f) The total number of ions in a mole of Cs3PO3 is:
i) 1
ii) 4
iii) 7
iv) 6.02 x 1023
v) 24.1 x 1023
vi) 42.1 x 1023
Verbal: Each FU contains four ions (three Cs+ ions and one PO33‐ ion), so a mole of the compound (6.022 x 1023 FU of it) would contain four moles of ions (four “moleion” ions). That equals 4 x (6.02 x 1023) ions or 24.1 x 1023 ions. ︶
2
3
︵
s
n s
o n
i
3 o
2 i
0
f
o
1 l
x o
2
2 m
0 1
.
6
x
s3
C
l
o
m
1
1 mol Cs3PO3 x
s O
n P
o
i
l
o
m
4
Calculational:  2.41 x 1024 ions (= 24.1 x 1023)
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2. Write T (True) or F (False) for the following statements. If the statement is false correct it
by changing, dropping, or adding a few words (2 points each)
(a) If reactants are combined in the ratio indicated by a chemical equation and reaction
occurs as completely as possible, then there will be no left over amount of either reactant.
TRUE
(b) If you calculate that 21 g of a product could form, but only 16 g of product are actually
obtained, then the percent yield was 31%. 76%.
FALSE
% yield =
actual yield
16 g
x 100 
x 100  76%
maximum (theoretical) yield
21 g
Free Response (x points in total)
3. (7 pts) Assign oxidation numbers to all of the atoms in all of the species in the following
chemical equation, and then state which species is the reducing agent (with reasoning).
2 I(aq)
+
S2O82-(aq)

I2(aq)
+
2 SO42-(aq)
Answers*: ‐1 +7 ‐2 0 +6 ‐2 * The oxidation numbers shown here below each atom type are for each atom of that type in the species. E.g., In S2O82‐, “‐16” is not an oxidation number for any atom; “‐2” is the oxidation # for each of the eight O’s in the second species. Answers with Reasoning: I‐ : Answer: ‐1 for I Reasoning: [my rule #3 on handout]: sum of ox #’s must equal net charge on species. Thus, if monotomic, the ox # = the actual charge. S2O82‐ : Answer: ‐2 for (each) O and +7 for (each) S Reasoning: Species is not an element. Assign each O ‐2 [my Rule 2]. Eight ‐2’s totals ‐16. 2x + (‐16) = ‐2 [Rule #3]  2x = +14  x = +7 (for S) I2 : Answer: 0 for (each) I Reasoning: [my rule #1 on handout]: Atoms in elements are assigned zero S2O82‐ : Answer: ‐2 for (each) O and +7 for S Reasoning: Species is not an element. Assign each O ‐2 (my Rule 2). four ‐2’s totals ‐8. x + (‐8) = ‐2 [Rule #3]  x = +6 (for S) ‐
Reducing agent:__ I __ Reasoning: (okay on an exam): A reducing agent gives electrons
away and thus its oxidation # becomes more positive. I’s ox #
went from -1 (in I-) to 0 (in I2) as reaction occurred showing that it
got more positive. Thus I- is the reducing agent.
(Full reasoning for those that may not understand how the reasoning above was determined):
The reducing agent is the one that “makes somebody else get reduced”, which means to “make somebody else’s (positive) charge get reduced. This means the other species must gain electrons, which means the reducing agent must lose them. If it loses electrons, it will get more positive. So look for the atom whose ox # gets more positive. Whichever reactant contains that atom is the reducing agent. Page 4
4. (6 pts) Write the net ionic equation. (Note: I have purposely left out some state designations)
La(NO3)3
+
3 HF
+

3 NH3(aq)
Answer: La3+ + 3 HF + 3 NH3(aq) 
LaF3(s) +
3 NH4NO3
LaF3(s) + 3 NH4+
Reasoning:
Only separate strong electrolytes (SEs) to get a complete ionic equation! Only two ways to be
a SE: 1) soluble ionic compound OR 2) one of the six strong acids.
La(NO3)3
SE
+
3 HF
not SE
+

3 NH3(aq)
not SE
LaF3(s) +
not SE
Not ionic, and not an acid HF is an acid, but it is not one of the six strong acids 3 NH4NO3
SE
Nitrates are soluble (Rule #1). [And NH4+ salts are soluble also!]. Soluble ionic compounds are strong electrolytes. Ionic, but not soluble. Nitrates are soluble (Rule #1). Soluble ionic compounds are strong electrolytes.
Thus: (NOTE: any species without a state designation below should have an “(aq)”. I left it out for clarity) Complete Ionic Eqn: La3+ + 3 NO3- + 3 HF + 3 NH3  LaF3(s) + 3 NH4+ + 3 NO3 Net Ionic Eqn: La3+ + 3 HF + 3 NH3 
LaF3(s) + 3 NH4+
5. (5 pts) Balance the following equation. (No work need be shown):
_4_PCl5 + _5_S4
__1_P4S10
+ _10__SCl2
I’d put the one first. Then put the 4 to balance the P’s (AVOID S here because it is in two places on the right side!). Then put the 10 to balance the Cl’s. Then balance the S’s last (don’t forget to include all the S’s on the right). 6. (10 pts)
C3H8 + 5 O2

3 CO2 + 4 H2O
(a) If 18.0 g of C3H8 and 70.0 g of O2 are put into a container, what is the maximum amount of
CO2, in moles, that could theoretically form assuming that the equation above represents the
reaction that takes place? You must show your work and reasoning
Answer: [For time reasons, I will have to leave this “shortened” for now; I hope to give a more complete and better formatted “answer” later]: C3H8 is limiting (0.409 moles C3H8 : 2.19 moles O2 PRESENT = 0.1868 : 1 < ratio of reaction
which is 1 : 5 = 0.2 : 1 => not enough C3H8. OR (alternate approach): Only 2.05 mol of O2
would react if all C3H8 reacts (0.409 x 5 = 2.05), and 2.19 moles are present, so O2 is in
excess.) Thus ALL of the C3H8 reacts. Thus, max. # moles of CO2 produced = 0.409 mol
C3H8 x 3/1 = 1.23 moles CO2.
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(b) List the chemical formulas of all substances that would be present in the container (in
theory) after the reaction in (a) is complete (must be consistent to answer in (a); no
explanation needed).
O2 (leftover), CO2 (product), and H2O (product) Although not asked, the amount of O2 left
over would be 2.19 mol – 2.05 mol = 0.14 mol.
7. (4 pts) Circle the member of each pair that contains a greater number of formula units of the
substance(s) described. Briefly indicate your reasoning.
(i) 58.9 g of Co or 58.9 g N2S
(ii) 1 g of calcium or 6.02 x 1023 calcium atoms
(iii) 1 molecule of O2 or 1 mol of O2
(iv) 1 mol of Cl or 1 mol of Cl2 SAME
Reasoning:
(i) 58.9 g of Co or 58.9 g Zn
58.9 g of Co has Avogadro’s number of FU’s (atoms) in it because 58.9 g of Co
(atoms) is one mole of Co atoms. One mole of N2S has a mass of about 60
g/mol, so 58.9 g of N2S (molecules) is LESS than one mole of N2S (molecules)
and so it would contain LESS than Avogadro’s number of FU’s (molecules here
since substance is molecular).
(ii) 1 g of calcium or 6.02 x 1023 calcium atoms
40.1 g of Ca is 1 mole of Ca (atoms), so 1 g of Ca is less than a mole of Ca atoms
(= 6.02 x 1023 Ca atoms).
(iii) 1 molecule of O2 or 1 mol of O2
(iv) 1 mol of Cl or 1 mol of Cl2
A mole of O2 is 6.02 x 1023 molecules so it obviously
has MUCH more than 1 molecule in it!
SAME! Each sample contains 6.02 x 1023 “things”; in
one case its atoms, in the other case its molecules.
8. (8 pts) (a) If you dilute 3.00 mL of a 0.0300 M KI solution to 10.00 mL, what will be the
concentration of KI in the final (diluted) solution?
This approach uses the fact that when one merely dilutes a solution, the amount of moles of solute 0.0300 M x (3.00 mL/10.00 mL) = 0.00900 M
remains the same. Just the volume increases. Thus, OR
“moles before”  “moles after”: MbVb  MaVa  Ma  Mb x Vb/Va
0.0300 mol/L x 0.00300 L  0.0000900 mol KI;
0.0000900 mol KI/0.01000 L = 0.00900 M
In this approach, the moles of solute are determined first from Mb and Vb, and then
the # moles is divided by the “after dilution” volume to determine molarity.
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(b) You wish to make up a solution of NaI that is 2.50 M in NaI using 60.0 g of NaI. What
volume, in mL, would your solution need to have in order to achieve your goal?
60.0 g / (149.9 g/mol)  0.400 mol NaI
0.400 mol NaI x (1 L / 2.50 mol)  0.160 L
0.160 L x 1000 mL/L  160. mL (or 1.60 x 102 mL)
9. (x pts) (a) Complete the equation for the following, assuming that a precipitate forms:
H2S
+

Zn(NO3)2
2 HNO3(aq) + ZnS(s)
Reasoning: Exchange reaction (ion swapping). H2S is thought of as being made up
of H+ and S2-; Zn(NO3)2 as Zn2+ and NO3-. Products are therefore nitric acid (H+ +
NO3- = HNO3 ) and zinc sulfide (Zn2+ + S2- = ZnS).
Reasoning:
H2S is not a SE b/c it is not
one of the six strong acids; it
is a weak acid.
(b) Write the net ionic equation that corresponds.

H2S
+
Zn(NO3)2
Complete ionic: H2S
+
Zn2+ + 2 NO3-
Net ionic: H2S
+
Zn2+

2 HNO3(aq) + ZnS(s)

Zn(NO3)2 is a SE b/c it is a
soluble ionic compound
(nitrate).
2 H+ + 2 NO3- + ZnS(s)
HNO3 is a SE b/c it is one of
the six strong acids.
2 H+ + ZnS(s)
ZnS is a solid because it must
be the precipitate (b/c HNO3 is
not); therefore it is insoluble.
(c) Draw nanoscopic pictures indicating what is asked for:
H+ =
S2- =
X
X
X
X
3 F.U. H2S
Zn2+ =
X
X
X
NO3- =
X
X
X
X
X
X
4 F.U. Zn(NO3)2
(Also, Rule #6 indicates sulfides
are generally insoluble & ZnS is
not an exception.)
X
X
X
X
X
mix
X
X
X X X
X
X
X
after mixing and reaction is complete
10. (8 pts) A certain sample of a compound containing phosphorus and sulfur was completely
decomposed to yield 21.92 g of phosphorus and 17.03 g of sulfur.
(a) (5 pts) Determine the empirical formula of the compound.
Answer: P4S3
Reasoning:
moles P 21.92 g / (30.974 g/mol P)  0.70769 mol P atoms
moles S  17.03 g / (32.066 g/mol S)  0.53109 mol S atoms
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Divide both by lower one (0.53109) to obtain ratio of P : S of 1.333 : 1
Multiply both by 3 to obtain ratio of P : S of 4 : 3
Write as formula: P4S3
(b) (3 pts) Calculate the mass percent of sulfur in the compound.
Answer: 43.72 % S (by mass)
Work/Reasoning:
mass of sulfur / total mass of sample  17.03 g S / (17.03 g S + 21.92 g P)  0.43722
Multiply by 100 (to convert to percent): 0.43722 x 100  43.72 %
[You could also use the formula from Part (a) to calculate the mass % of S; the answer
will vary ever so slightly:
[3 x 32.07 / (4 x 30.97 + 3 x 32.07)] x 100  43.71 % ]
11. (5 pts) What is the mass of 0.632 moles of Pb(C2O4)2?
Answer: 242 g
Work/Reasoning:
Molar mass of Pb(C2O4)2  207.2 + 4(12.01) + 8(16.00) = 383.24 g/mol
So 0.632 moles of the compound would have a mass of
0.632 mol x 383.24 g/mol)  242.2 = 242 g
12. (x pts)
5 C(s) + 2 SO2(g)

CS2(l) + 4 CO(g)
(a) How many moles of CS2 are formed if 0.832 moles of C react according to the equation.
0.832 mol C react x 1 mol CS2 form / 5 mol C that react  0.166 mol CS2 form.
(b) How many grams of C need to react to make 4.68 grams of CO according to the equation.
4.68 g CO x 1 mol CO / 28.01 g  0.1671 mol CO form.
0.1671 mol CO form x 5 mol C react / 4 mol CO that form  0.2089 mol C react
0.2089 mol C x 12.01 g/mol  2.508  2.51 g C needed (to react)
OR all in one “big” (3-step) calculation:
4.68 g CO x [1 mol CO / 28.01] x [5 mol C / 4 mol CO] x [12.01 g / 1 mol C]  2.51 g C
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13. (8 pts) (a) How many moles of Fe are in 34.6 g of Fe?
“Conceptual” way: 55.845 g Fe is one mole, so 34.6 g is 34.6/55.845 ths of a mole or 0.61957 mol =
0.620 mol
OR using full units:
34.6 g Fe
 0.620 mol Fe
55.845 g Fe/mol Fe
OR
34.6 g Fe x 1 mol / 55.845 g = 0.620 mol Fe
(b) What is the mass of 0.632 moles of Pb?
“Conceptual” way: If one mole of Pb has a mass of 207.2 g, then 0.632 moles is 0.632 x 207.2 g =
130.95 g = 131 g
 207.2 g Pb 
  0.632 mol Pb   131 g Pb
 1 mol Pb 
OR using full units: 
(c) What is the mass of 2.5 x 1013 moles of N?
“Conceptual” way : If each mole of N atoms has a mass of 14.0 g, then 2.5 x 1013 moles will have
2.5 x 1013 times as much mass as 14.0 g, or (2.5 x 1013)(14.0 g ) = 3.5 x 1014 g
 14.0 g N 
2.5 x 1013 mol N  3.5 x 1014 g N

 1 mol N 

OR using full units: 

(d) What is the mass of 2.5 x 1013 atoms of N?
“Conceptual” way: 6.02 x 1023 atoms of N is a mole, so this sample is MUCH less than a mole. In
fact it is:
(2.5 x 1013)/(6.02 x 1023)ths of a mole, which is 4.153 x 10-11 moles.
But each mole has a mass of 14.0 g, so the mass of 4.153 x 10-11 moles is just
4.153 x 10-11 x 14.0 g = 5.81 x 10-10 g = 5.8 x 10-10 g


2.5 x 1013 N atoms
-11
  4.153 x 10 mol N;
23
6.02
x
10
N
atoms/mol
of
N
atoms


OR using full units: 
[also could set up as:
 4.153 x 10
-11
2.5 x 1013 atoms x 1 mol / 6.02 x 1023 atoms = 4.153 x 10-11 mol]
 14.0 g N 
-10
-10
mol N 
  5.814 x 10 g N = 5.8 x 10 g N
mol
N


