Announcements
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Exam tonight, 7-8:15pm (locations
posted on website)
Conflict Exam, 5:15-6:30pm (114
Transportation Bldg)
No lab this week!
Start new material on Thursday (read
chapter 10!)
Review Questions (Exam II)
1. b
2. b
3. e
4. e
5. b
6. d
7. e
8. c
9. b
10. a
11. c
12. c
13.
14.
15.
16.
17.
18.
19.
20.
c
e
b
a
d
c
a
c
22. a) ethanol = LR
0.0870 mol CO2
0.131 mol H2O
b) 6.57 L
c) 12.0g = 12.0g
Fall 2010 Exam II
1.
2.
3.
4.
5.
6.
7.
e
b
d
d
c
c
a
8. e
9.
10.
11.
12.
13.
14.
15.
a
c
e
a
b
c
b
16. a) same
b) Soln #1
c) 6.38 M
d) 45.5 mL
17. b) 37.4 g NF3
c) 40. g = 40. g
d) smaller than
Hour Exam II, Fall 2010
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
1) 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
How many moles of steam must react to produce
1.62 moles of Fe3O4?
1.62 mol Fe3O4 * 4 mol H2O = 6.48 moles H2O
1 mol Fe3O4
Answer is (e).
Hour Exam II, Fall 2010

2) How many of the following statements about
chemical reactions is/are false?


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I. When balancing a chemical equation, all subscripts must
be conserved.
II. When one coefficient is doubled, the rest of the
coefficients in the balanced equation must also be doubled.
III. An individual coefficient in a balanced equation is
meaningless.
IV. The phases in a chemical reaction tell us the nature of
the reactants and products.
Answer is (b).
Hour Exam II, Fall 2010
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
3) 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
Suppose 2.75 moles of HNO3 were produced from 30.0 g of
water and a certain amount of nitrogen dioxide. Which of the
following statements is true?

a) Water is the limiting reactant because 30.0 g of water
produces 2.75 mol HNO3.
 b) Water is the limiting reactant because it has a smaller
coefficient than the nitrogen dioxide.
 c) Water is not the limiting reactant because it has a smaller
coefficient than the nitrogen dioxide.
 d) Water is not the limiting reactant because more than
2.75 mol HNO3 could be produced from 30.0 g of water.
 e) There is not enough information given to answer this
question.
Hour Exam II, Fall 2010
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3) 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
Suppose 2.75 moles of HNO3 were produced from 30.0 g of
water and a certain amount of nitrogen dioxide. Which of the
following statements is true?
How many moles is 30.0 g of water?
 30.0 g H2O * (mol/18.016g) = 1.67 moles H2O
How many moles of HNO3 could I make from that?
 1.67 moles H2O * (2 mol HNO3/1 mol H2O) = 3.34 moles
HNO3
Since you can possibly make 3.34 moles of HNO3 from 30.0 g
water, but the reaction only made 2.75 moles HNO3, water
cannot be limiting!
Answer is (d).
Hour Exam II, Fall 2010
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4) How many moles of potassium ions are in 80.0
mL of a 2.00M potassium sulfide solution?
potassium sulfide: K+ and S2- so formula is K2S
Moles of K2S:
80.0 mL * 1L/1000mL * 2.00 moles/L = 0.16 mol K2S
Moles of K+:
0.16 moles K2S * 2 mol K+ = 0.32 moles K+
1 mol K2S
Answer is (d).
Hour Exam II, Fall 2010
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5) 2NaN3(s)  2Na(s) + 3N2(g)
What mass of NaN3 must be used to inflate an air bag to 105L
at 1.00 atm and 25C?
Moles of N2 in air bag: PV = nRT; n = PV/RT
n = (1.00 atm)(105L) = 4.2938 mol N2
(0.08206)(25+273)
Moles NaN3 needed to make N2:
4.2938 mol N2 * 2 mol NaN3 = 2.8625 mol NaN3
3 mol N2
Mass of NaN3: 2.8625 mol NaN3 * (65.01g/mol) = 186 g NaN3
Answer is (c).
Hour Exam II, Fall 2010
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6) Fe2O3(s) + X(s)  Fe(l) + X2O3(s)
Balance the equation in standard form and determine
the sum of the coefficients.
Fe2O3(s) + 2X(s)  2Fe(l) + X2O3(s)

Sum of coefficients is 6.

Answer is (c).
Hour Exam II, Fall 2010
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7) Fe2O3(s) + 2X(s)  2Fe(l) + X2O3(s)
If 77.7 g of X reacts with excess iron(III) oxide to produce 1.44
moles of X2O3, what is the identity of X?
Need to determine molar mass of X (g/mol)
Grams of X: 77.7 g
Moles of X:
1.44 moles X2O3 * 2 mol X
= 2.88 mol X
1 mol X2O3
So molar mass would be:
77.7 g X = 26.98 g/mol, which is aluminum
2.88 mol X
Answer is (c).
Hour Exam II, Fall 2010

8) Consider the equation: 3A + B  2C. The molar mass of A is
40.0 g/mol. Which of the following statements is true when
equal masses of A and B are reacted?
a) If the molar mass of B is greater than the molar mass of A, then B must
determine how much C is produced.
b) If the molar mass of B is less than the molar mass of A, then B must
determine how much C is produced.
c) If the molar mass of B is the same as the molar mass of A, then A and B
react in a perfect stoichiometric ratio and both determine how much C is
produced.
d) If the molar mass of B is greater than the molar mass of A, then A must
determine how much C is produced.
e) If the molar mass of B is less than the molar mass of A, then A must
determine how much C is produced.
Hour Exam II, Fall 2010


8) Consider the equation: 3A + B  2C. The molar mass of A is
40.0 g/mol. Which of the following statements is true when
equal masses of A and B are reacted?
Assume 40 g of A and B, so A is 1 mole. So moles of B needed to react with 1
mole of A:
1 mol A * (1 mol B/3 mol A) = 0.33 mol B needed
If the molar mass of B is greater than the molar mass of A:
40g B * (mol/50g) = 0.8 mol B; A is limiting
40g B * (mol/200g) = 0.2 mol B; B is limiting
If the molar mass of B is less than the molar mass of A:
40g B * (mol/30g) = 1.33 mol B; A is limiting
40g B * (mol/10g) = 4.0 mol B; A is limiting
If the molar mass of B is the same as the molar mass of A:
40g B * (mol/40g) = 1.0 mol B; A is limiting
Hour Exam II, Fall 2010
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9)
Na2SiF6(s) + Na(s)  Si(s) + NaF(s)
Balanced equation:
Na2SiF6(s) + 4Na(s)  Si(s) + 6NaF(s)
If 3.75 moles of NaF were produced, how many
moles of Si were also created?
3.75 mol NaF * 1 mol Si = 0.625 mol Si
6 mol NaF
Answer is (a).
Hour Exam II, Fall 2010
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10) Na2SiF6(s) + 4Na(s)  Si(s) + 6NaF(s)
How many grams of Na2SiF6 (MW = 188.07 g/mol) are required
to completely react with 0.555 mol Na?
Use moles of Na and ratios of coefficients to find moles of
Na2SiF6:
0.555 mol Na * 1 mol Na2SiF6 = 0.13875 mol Na2SiF6
4 mol Na
Then use molar mass to find grams:
0.1378 mol Na2SiF6 * (188.07g/mol) = 26.1 g Na2SiF6
Answer is (c).
Hour Exam II, Fall 2010

11)
Co2+
ClNa+
OH-





What is in beaker #1? 4 mol Co2+ and 8 mol ClMolecular formula of compound in beaker #1: CoCl2
What is in beaker #2? 4 mol Na+ and 4 mol OHMolecular formula of compound in beaker #2: NaOH
What products would form in this reaction? Co(OH)2 and NaCl
Hour Exam II, Fall 2010

11)
Co2+
ClNa+
OH-





Unbalanced equation for reaction:
CoCl2(aq) + NaOH(aq)  Co(OH)2 + NaCl
Which one is the precipitate? Co(OH)2
Balanced equation:
CoCl2(aq) + 2NaOH(aq)  Co(OH)2(s) + 2NaCl(aq)
Hour Exam II, Fall 2010

11)
Co2+
ClNa+
OH-





Balanced equation:
CoCl2(aq) + 2NaOH(aq)  Co(OH)2(s) + 2NaCl(aq)
Complete ionic equation:
Co2+(aq) + 2Cl-(aq) + 2Na+(aq) + 2OH-(aq)  Co(OH)2(s) +
2Na+(aq) + 2Cl-(aq)
Answer is (e).
Hour Exam II, Fall 2010

12)
Co2+
ClNa+
OH-

Which of the ions are in solution after the given reaction takes
place?
Hour Exam II, Fall 2010

12)
Co2+
ClNa+
OH-
Hour Exam II, Fall 2010

12) Which of the following ions are in solution after
the given reaction takes place?
I. Na+
II. OHIII. Co2+
IV. Cl-

I, III, and IV

Answer is (a).




Co2+
Cl-
Na+
OH-
Hour Exam II, Fall 2010
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13) 10.0 mL of a 0.30M sodium chromate solution
reacts with 20.0 mL of a 0.20M aluminum bromide
solution. A precipitate forms.
What are the reactants?
Na+ and CrO42- so formula is Na2CrO4
Al3+ and Br- so formula is AlBr3
What are the products?
Na+ and Br- so formula is NaBr
Al3+ and CrO42- so formula is Al2(CrO4)3
Hour Exam II, Fall 2010
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13) 10.0 mL of a 0.30M sodium chromate solution
reacts with 20.0 mL of a 0.20M aluminum bromide
solution. A precipitate forms.
Unbalanced equation:
Na2CrO4(aq) + AlBr3(aq)  NaBr + Al2(CrO4)3
Balanced equation:
3Na2CrO4(aq) + 2AlBr3(aq)  6NaBr + Al2(CrO4)3
Which one is the precipitate?
NaBr is soluble according to rules so Al2(CrO4)3 is the
solid formed
Hour Exam II, Fall 2010
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13)
Balanced equation:
3Na2CrO4(aq) + 2AlBr3(aq)  6NaBr(aq) + Al2(CrO4)3(s)
Moles of reactants:
10.0 mL * (L/1000mL) * 0.30 mol/L = 0.003 mol Na2CrO4
20.0 mL * (L/1000mL) * 0.20 mol/L = 0.004 mol AlBr3
Limiting reactant:
0.003 mol Na2CrO4 * 2 mol AlBr3 = 0.002 mol AlBr3 needed
3 mol Na2CrO4
Have 0.004 mol AlBr3 (only need 0.002) so Na2CrO4 is limiting
reactant
Hour Exam II, Fall 2010

13)
Balanced equation:

3Na2CrO4(aq) + 2AlBr3(aq)  6NaBr(aq) + Al2(CrO4)3(s)
 Moles of product:
0.003 mol Na2CrO4 * 1 mol Al2(CrO4)3 = 0.001 mol Al2(CrO4)3
3 mol Na2CrO4

Mass of product:
0.001 mol Al2(CrO4)3 * 401.96g Al2(CrO4)3/mol = 0.40 g Al2(CrO4)3

Answer is (b).
Hour Exam II, Fall 2010
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14)
3Na2CrO4(aq) + 2AlBr3(aq)  6NaBr(aq) + Al2(CrO4)3(s)
What is the concentration of sodium ions left in solution after
the reaction is complete?
Complete ionic equation:
+
23+(aq) + 6Br-(aq)  6Na+(aq)
 6Na (aq) + 3CrO4 (aq) + 2Al
+ 6Br-(aq) + Al2(CrO4)3(s)
Net ionic equation:
3+(aq) + 3CrO 2-(aq)  Al (CrO ) (s)
 2Al
4
2
4 3
Na+ does not participate in reaction so:
moles Na+ in solution before reaction = moles Na+ in solution
after reaction
Hour Exam II, Fall 2010
14)
0.003 mol Na2CrO4 * 2 mol Na+ = 0.006 mol Na+
1 mol Na2CrO4
Final volume: 10.0 + 20.0 = 30.0 mL = 0.030 L

Molarity = 0.006 mol Na+ = 0.20M Na+
0.030 L
Answer is (c).
Hour Exam II, Fall 2010


15)
Net ionic equation:
3+(aq) + 3CrO 2-(aq)  Al (CrO ) (s)
 2Al
4
2
4 3
0.003 mol Na2CrO4 (limiting) and 0.004 mol AlBr3
 Moles of Al3+ before reaction:
0.004 mol AlBr3 * 1 mol Al3+ = 0.004 mol Al3+
1 mol AlBr3
2 Moles of CrO4 :
0.003 mol Na2CrO4 * 1mol CrO42- = 0.003 mol CrO421 mol Na2CrO4

Hour Exam II, Fall 2010
Moles of Al3+ used:
0.003 mol CrO42- * 2 mol Al3+ = 0.002 mol Al3+ used
3 mol CrO42 Moles of Al3+ left:
3+
 0.004 – 0.002 = 0.002 mol Al
 Concentration:
0.002 mol Al3+= 0.067 M Al3+
0.030 L
 Answer is (b).

Hour Exam II, Fall 2010
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

16a)
You have three magnesium nitrate solutions on a lab table in front of
you. All of the solutions came from a 500.0 mL volumetric flask
containing 3.00 M Mg(NO3)2.
How does the concentration of Solution #1 compare to the
concentration of Solution #2? Justify your answer.

Solution #1 and Solution #2 have the same concentration (3.00 M)
since both solutions came from the same flask (there is still the
same ratio of moles/L in each solution).
Hour Exam II, Fall 2010
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16b)
Which solution (not including the volumetric flask) contains the
greatest number of ions? Provide complete mathematical support in
your answer below.
Solution #1: 100.0 mL of 3.00 M Mg(NO3)2
100.0 mL * L/1000mL * 3.00 mol/L * 3 mol ions/1 mol Mg(NO3)2 =
0.900 mol ions
Solution #2: 50.0 mL of 3.00 M Mg(NO3)2
50.0 mL * L/1000mL * 3.00 mol/L * 3 mol ions/1 mol Mg(NO3)2 =
0.450 mol ions
Solution #3: 10.0 mL of 3.00 M Mg(NO3)2
10.0 mL * L/1000mL * 3.00 mol/L * 3 mol ions/1 mol Mg(NO3)2 =
0.0900 mol ions
Solution #1 has the greatest number of ions (0.900 mol)
Hour Exam II, Fall 2010


16c)
What is the new concentration of Solution #2 if 25.0 g of solid
magnesium nitrate (MW = 148.33 g/mol) is added and dissolved?
(Assume no volume change.)
25.0 g Mg(NO3)2 * 1 mol Mg(NO3)2 = 0.169 mol Mg(NO3)2 added
148.33 g Mg(NO3)2
0.169 + 0.150 = 0.319 mol total
M = 0.319 mol = 6.38 M Mg(NO3)2
0.050 L
Hour Exam II, Fall 2010

16d)
What volume of water (in mL) must evaporate from Solution #1 in
order to have a concentration of 5.50 M?
M1V1 = M2V2

(3.00 M)(0.100 L) = (5.50 M) * V2

V2 = 0.0545 L or 54.5 mL

100.0 – 54.5 mL = 45.5 mL of water must evaporate


Hour Exam II, Fall 2010
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
17a) N2(g) + 3F2(g)  2NF3(g)
Assume this reaction takes place in an elastic balloon with an
atmospheric pressure of 1.00 atm and a temperature of 25C.
Scenario I: You have a stoichiometric mixture of nitrogen and fluorine
gases (neither reactant is limiting). Draw a molecular-level diagram
before and after the reaction occurs, including the relative sizes of the
balloons.
Hour Exam II, Fall 2010
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17b) N2(g) + 3F2(g)  2NF3(g)
Assume this reaction takes place in an elastic balloon with an
atmospheric pressure of 1.00 atm and a temperature of 25C.
Scenario II: You react 10.0 g nitrogen gas with 30.0 g of fluorine gas
at constant temperature and pressure.
Determine the mass of NF3 that can be produced from these two
reactants.
Moles of N2: 10.0g * mol/28.02g = 0.357 mol N2
Moles of F2: 30.0g * mol/38.00g = 0.789 mol F2
Limiting reactant:
0.357 mol N2 * 3 mol F2/1 mol N2 = 1.071 mol F2 needed
Only have 0.789 mol so F2 is limiting
Hour Exam II, Fall 2010
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
17b) N2(g) + 3F2(g)  2NF3(g)
Assume this reaction takes place in an elastic balloon with an
atmospheric pressure of 1.00 atm and a temperature of 25C.
Scenario II: You react 10.0 g nitrogen gas with 30.0 g of fluorine gas
at constant temperature and pressure.
Determine the mass of NF3 that can be produced from these two
reactants.
Moles of product:
0.789 mol F2 * 2 mol NF3/3 mol F2 = 0.526 mol NF3
Mass of product:
0.526 mol NF3 * 71.01 g/mol = 37.4 g NF3
Hour Exam II, Fall 2010
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
17c) N2(g) + 3F2(g)  2NF3(g)
Assume this reaction takes place in an elastic balloon with an
atmospheric pressure of 1.00 atm and a temperature of 25C.
Scenario II: You react 10.0 g nitrogen gas with 30.0 g of fluorine gas
at constant temperature and pressure.
Verify that mass has been conserved for Scenario II within two
significant figures.
Mass before the reaction: 10.0 + 30.0 = 40.0 g total
Mass of N2 left over:
0.789 mol F2 * 1 mol N2/3 mol F2 = 0.263 mol N2 used
0.357 – 0.263 = 0.0939 mol N2 left
0.0939 mol N2 * 28.01 g/mol = 2.63 g N2 left
Mass after the reaction: 37.4 g NF3 + 2.63 g N2 = 40.0 g total
Hour Exam II, Fall 2010
17d) N2(g) + 3F2(g)  2NF3(g)

Assume this reaction takes place in an elastic balloon with an
atmospheric pressure of 1.00 atm and a temperature of 25C.

For Scenario II, the volume of the balloon after the reaction takes
place will be ________ the volume of the balloon before the reaction
took place.

PV = nRT; at constant P and T:
n1 = n2
V1 V2
n1 (before): 0.357 mol N2 + 0.789 mol F2 = 1.146 mol total
n2 (after): 0.526 mol NF3 + 0.0939 mol N2 = 0.6202 mol total

Since there is a direct relationship between n and V, and n has decreased,
the volume will also decrease, so the balloon will be SMALLER.
Fall 2011 Exam II
1.
2.
3.
4.
5.
6.
7.
8.
d
e
b
a
b
b
c
d
9.
10.
11.
12.
13.
14.
15.
e
c
d
c
a
c
a
16. c)
Take 48.1 mL of 2.00 M
NaCl; add water until you
get 275 mL total soln
17.
a)
i) Beaker D
ii) Beaker B
iii) Beaker C
iv) Beaker A
b) 10.0 g PbCO3
c) i) Yes; PbCl2
ii) NO3–: 0.667 M
K+: 0.500 M
Na+: 0.0833 M
Hour Exam II, Fall 2011

1) What information can we obtain from a balanced
equation such as 2H2(g) + O2(g)  2H2O(g)? Choose
the best answer.
a) The formula of each reactant and product
b) The phase of each reactant and product
c) The starting amount of each reactant
d) Both a and b are true
e) All of the above (a-c) are true
Hour Exam II, Fall 2011

2) Consider the following unbalanced equation:


Balanced equation:


C2H6(g) + 7/2 O2(g)  2CO2(g) + 3H2O(g)
Standard form:


C2H6(g) + O2(g)  CO2(g) + H2O(g)
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
How many of the following represent a correct mole ratio
between C2H6 and O2 when the equation is balanced?
I.
II.
III.
IV.

1
2
4
8
:
:
:
:
7/2
7
14
28
Answer is (e).
Hour Exam II, Fall 2011

3)Fe2O3(s) + 2Al(s)  2Fe(l) + Al2O3(s)
What mass of iron(III) oxide must be used to
produce 15.0 g of iron?
15.0 g Fe * mol/55.85g = 0.2689 mol Fe
0.2689 mol Fe * 1 mol Fe2O3 = 0.1343 mol Fe2O3
2 mol Fe
0.1343 mol Fe2O3 * 159.7 g/mol = 21.4 g Fe2O3

Answer is (b).





Hour Exam II, Fall 2011

4)Fe2O3(s) + 2Al(s)  2Fe(l) + Al2O3(s)
When 15.0 g of iron is produced, what is the
maximum mass of aluminum oxide created?
0.2689 mol Fe * 1 mol Al2O3 = 0.1343 mol Al2O3
2 mol Fe
0.1343 mol Al2O3 * 101.96 g/mol = 13.7 g Al2O3

Answer is (a).




Hour Exam II, Fall 2011
5) C(s) + O2(g)  CO2(g)
 What volume of oxygen gas at 30.C and 1.20 atm
would be required to react completely with 1.00 g of
carbon?
 1.00 g C * mol/12.01g = 0.0834 mol C
 0.0834 mol C * 1 mol O2 = 0.0834 mol O2
 PV = nRT; V = nRT/P
V = (0.0834)(0.08206)(303) = 1.73 L
1.20


Answer is (b).
Hour Exam II, Fall 2011


6) In which of the following situations would a chemical
reaction not take place when the two solutions are mixed?
a) AgNO3(aq) is mixed with NH4Cl(aq)


b) Na2CO3(aq) is mixed with K2S(aq)


BaSO4 and NaNO3; BaSO4 is a solid
d) HCl(aq) is mixed with NaOH(aq)


Na2S and K2CO3; both are soluble, no net ionic equation
c) Ba(NO3)2(aq) is mixed with Na2SO4(aq)


AgCl and NH4NO3; AgCl is a solid
NaCl and H2O; no solid, but can write a net ionic equation!
e) a chemical reaction would take place in all of the situations
above
Hour Exam II, Fall 2011
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7) Unbalanced equation:
NH3 + O2  NO + H2O
For every 1.00 mol of NH3 that reacts, ____ mol of O2 is
required.
Balanced equation:
2NH3 + 5/2 O2  2NO + 3H2O
Standard form:
4NH3 + 5O2  4NO + 3H2O
For 1.00 mol NH3:
1.00 mol NH3 * 5 mol O2 = 1.25 mol O2 required
4 mol NH3
Answer is (c).
Hour Exam II, Fall 2011
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8) A 10.50 g sample of magnesium chloride is measured and
weighed. The solid sample is dissolved in water. Enough water
is added to make a 500.0 mL solution.
What is the concentration of magnesium chloride in the
solution?
10.50 g MgCl2 * mol/95.21 g = 0.1103 mol MgCl2
0.1103 mol MgCl2 = 0.2206 M MgCl2
0.5000 L
Answer is (d).
Hour Exam II, Fall 2011
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9)
How many chloride ions are in the solution?
0.1103 mol MgCl2 * 2 mol Cl- = 0.2206 mol Cl1 mol MgCl2
0.2206 mol Cl- * 6.022 x 1023 ions = 1.328 x 1023 Cl- ions
1 mol
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Answer is (e).
Hour Exam II, Fall 2011
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10) Consider the following representation of a
chemical equation:
Balance the equation in standard form and determine
the sum of the coefficients.
Hour Exam II, Fall 2011
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10) Consider the following representation of a
chemical equation:
2

+
Answer is (c).
1
+
2
= 5
Hour Exam II, Fall 2011
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11) Consider the equation: 4A + 3X2  2A2X3. Which
of the following must be conserved?
I. Mass
II. Moles
III. Subscripts
IV. Number of atoms

Answer is (d).
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Hour Exam II, Fall 2011
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12) You add 200.0 mL of water to 200.0 mL of a 0.50
M sugar solution. Which of the following will not
change?
a) total volume of the solution
b) concentration of the solution
c) moles of solute of the solution
d) total mass of the entire solution
e) both c and d will not change
Hour Exam II, Fall 2011
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13)
2C6H5Cl
+ C2HOCl3  C14H9Cl5 + H2O
Name:
chlorobenzene
chloral
DDT

MW:
112.55
147.38
354.46

In a government lab, 1150 g of chlorobenzene is reacted with 975 g of
chloral. What mass of DDT is formed?
1150 g C6H5Cl * mol/112.55g = 10.22 mol C6H5Cl
975 g chloral * mol/147.38g = 6.62 mol chloral

Chloral needed to react:
10.22 mol C6H5Cl * 1 mol chloral = 5.11 mol chloral  chloral in excess
2 mol C6H5Cl

Mass of DDT:
10.22 mol C6H5Cl * 1 mol DDT * 354.46 g/mol = 1.81 x 103 g DDT
2 mol C6H5Cl

Answer is (a).
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Hour Exam II, Fall 2011
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14) What mass of excess reactant is left over?
Chloral started with:
6.62 mol chloral
Chloral needed:
- 5.11 mol chloral
1.51 mol chloral left over
1.51 mol chloral * 147.38 g/mol = 222 g chloral

Answer is (c).
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Hour Exam II, Fall 2011
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a)
b)
c)
d)
e)
15) As NH3(g) is decomposed into nitrogen gas and hydrogen gas at
constant pressure and temperature, how does the volume of the
product gases collected compare to the volume of NH3 reacted? Choose
the best answer.
2NH3(g)  N2(g) + 3H2(g)
The volume of the product gases is twice the volume of NH3
reacted because twice the number of moles are produced.
The volume of the product gases is half the volume of NH3 reacted
because the molecules become smaller in size.
The volume of the product gases is the same as the volume of NH3
reacted because the pressure is constant.
The volume of the product gases is the same as the volume of NH3
reacted because the law of conservation of mass is followed.
The volumes cannot be compared to one another without actual data
given.
Hour Exam II, Fall 2011
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16a) Write and balance the reactions below in
standard form, including all phases:
(i) Solid magnesium metal reacts with solid
manganese(III) oxide to produce solid magnesium
oxide and solid manganese metal.
3Mg(s) + Mn2O3(s) -> 3MgO(s) + 2Mn(s)
Hour Exam II, Fall 2011
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16a) Write and balance the reactions below in
standard form, including all phases:
(ii) Xenon gas will combine directly with fluorine gas
to produce solid xenon tetrafluoride.
Xe(g) + 2F2(g) -> XeF4(s)
Hour Exam II, Fall 2011
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16a) Write and balance the reactions below in
standard form, including all phases:
(iii) Aqueous potassium hydroxide is mixed with
aqueous nickel(II) chloride. Write the balanced
molecular, complete ionic, and net ionic equations in
standard form and include all phases:
Reactants:
KOH and NiCl2
Products:
Ni(OH)2 and KCl
Hour Exam II, Fall 2011
16a) Write and balance the reactions below in
standard form, including all phases:
 (iii) Balanced molecular equation:
2KOH(aq) + NiCl2(aq) -> Ni(OH)2(s) + 2KCl(aq)
 Complete ionic equation:
2K+(aq) + 2OH-(aq) + Ni2+(aq) + 2Cl-(aq) ->
Ni(OH)2(s) + 2K+(aq) + 2Cl-(aq)
 Net ionic equation:
2OH-(aq) + Ni2+(aq) -> Ni(OH)2(s)

Hour Exam II, Fall 2011
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16b) what is a limiting reactant and why is it
important in chemistry? Please limit your answer to
the space provided.
The limiting reactant is consumed (used up) in the
reaction. This is important in chemistry because the
limiting reactant determines the amount of product
that is made.
Hour Exam II, Fall 2011
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16c) How do you prepare 275 mL of a 0.350 M NaCl
solution using an available 2.00 M solution? Support
your answer with calculations.
M1V1 = M2V2
(0.275L)(0.350M) = (x)(2.00M)
x = 0.048125L = 48.1 mL
Take 48.1 mL of a 2.00 M NaCl solution and add
water until you get 275 mL of total solution (add
226.9 mL of water).
Hour Exam II, Fall 2011
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
17a)
i) lead(II) nitrate – beaker D. The formula is Pb(NO3)2, which
dissociates to Pb2+ and 2NO3-. This represents a 1:2 ratio of a 2+
charge to a 1- charge.
ii) sodium chloride – beaker B. The formula is NaCl, which dissociates
to Na+ and Cl-. This represents a 1:1 ratio of a 1+ charge to a 1charge.
iii) potassium carbonate – beaker C. The formula is K2CO3, which
dissociates to 2K+ and CO32-. This represents a 2:1 ratio of a 1+
charge to a 2- charge.
iv) magnesium sulfate – beaker A. The formula is MgSO4, which
dissociates to Mg2+ and SO42-. This represents a 1:1 ratio of a 2+
charge to a 2- charge.
Hour Exam II, Fall 2011
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17b)
If 50.0 mL of a 1.00 M lead(II) nitrate solution is mixed with 75.0 mL
of a 0.500 M potassium carbonate solution, what mass of precipitate
will form?

Pb(NO3)2(aq) + K2CO3(aq)  PbCO3(s) + 2KNO3(aq)

Moles of reactants:
50.0 mL * 1L/1000mL * 1.00 mol Pb(NO3)2/L = 0.0500 mol Pb(NO3)2
75.0 mL * 1L/1000mL * 0.500 mol K2CO3/L = 0.0375 mol K2CO3

Limiting reactant:
0.0500 mol Pb(NO3)2 * 1 mol K2CO3 = 0.0500 mol K2CO3
1 mol Pb(NO3)2
K2CO3 is limiting
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Hour Exam II, Fall 2011
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17b)
If 50.0 mL of a 1.00 M lead(II) nitrate solution is mixed with 75.0 mL
of a 0.500 M potassium carbonate solution, what mass of precipitate
will form?
Pb(NO3)2(aq) + K2CO3(aq)  PbCO3(s) + 2KNO3(aq)
K2CO3 is limiting
Mass of product:
0.0375 mol K2CO3 * 1 mol PbCO3 * 267.21 g PbCO3 = 10.0 g PbCO3
1 mol K2CO3
mol
Hour Exam II, Fall 2011
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17c) After the reaction occurs in part b, 25.0 mL of a
0.500M sodium chloride solution is added.
i) Does a precipitate form? Why or why not? Justify
your answer.
Yes. Pb2+ is in excess from part b. Pb2+ reacts with
the Cl- to form solid PbCl2.
Hour Exam II, Fall 2011
17c) After the reaction occurs in part b, 25.0 mL of a
0.500M sodium chloride solution is added.
 i) What is the concentration of each spectator ion
after all of the reactions are complete? (assume the
volumes are additive).
 Spectator ions are K+, NO3-, and Na+.
 K +:
0.0375 mol K2CO3 * 2 mol K+ = 0.0750 mol K+
1 mol K2CO3
0.0750 mol = 0.500 M K+
0.150L

Hour Exam II, Fall 2011
17c) After the reaction occurs in part b, 25.0 mL of a
0.500M sodium chloride solution is added.
 i) What is the concentration of each spectator ion
after all of the reactions are complete? (assume the
volumes are additive).
 NO3-:
0.0500 mol Pb(NO3)2 * 2 mol NO3- = 0.100 mol NO31 mol Pb(NO3)2
0.100 mol = 0.667 M NO30.150L

Hour Exam II, Fall 2011
17c) After the reaction occurs in part b, 25.0 mL of a
0.500M sodium chloride solution is added.
 i) What is the concentration of each spectator ion
after all of the reactions are complete? (assume the
volumes are additive).
 Na+:
25.0mL * L/1000mL * 0.500 mol/L = 0.0125 mol NaCl
0.0125 mol NaCl * 1 mol Na+ = 0.0125 mol Na+
1 mol NaCl
0.0125 mol = 0.0833 M Na+
0.150L
