CHEMISTRY 101
Hour Exam II
October 31, 2006
Adams/Le
Name __________KEY_____________________
Signature ________________________________
T.A./Section ______________________________________
“Research tells us fourteen out of any ten individuals like chocolate.”
Happy Halloween!
This exam contains 17 questions on 6 numbered pages. Check now to make sure you have a
complete exam. You have one hour and fifteen minutes to complete the exam. Determine
the best answer to the first 15 questions and enter these on the special answer sheet. Also,
circle your responses in this exam booklet. Show all of your work and provide complete
answers to questions 16 and 17.
1-15
(30 pts.)
_________
16
(14 pts.)
_________
17
(16 pts.)
_________
(60 pts)
_________
Total
Useful Information:
1.000 L = 1000.0 mL
Always assume ideal behavior for gases (unless explicitly told otherwise).
PV = nRT
R = 0.08206 L·atm/mol·K
K = °C + 273
NA = 6.022 x 1023
Solubility Rules:
1. Most nitrate salts are soluble.
2. Most salts of sodium, potassium, and ammonium cations are soluble.
3. Most chloride salts are soluble. Exceptions: silver(I), lead(II), and mercury(I) chloride.
4. Most sulfate salts are soluble. Exceptions: calcium, barium, and lead (II) sulfate.
5. Most hydroxide salts can be considered insoluble. Soluble ones: sodium, potassium, and
calcium hydroxide.
6. Consider sulfide, carbonate, and phosphate salts to be insoluble. Soluble ones: sodium
and potassium.
CHEMISTRY 101
Hour Exam II
Fall 2006
Page No. 1
1. Balance the following equation in standard form and determine the sum of the
coefficients.
Na 2S2 O3 (aq) + I 2 (aq) →
a) 4
b) 5
Na 2S4 O6 (aq) + NaI(aq)
c) 6
d) 8
e) 9
2. You have equal concentrations of different solutes in equal volumes of solution. Which
of these solutions contains the least mass of solute?
a) NH4Cl
b) CaCl2
c) MgCl2
d) SrCl2
e) All the same.
---------------------------------------------------------------------------------------------------------------Use the following unbalanced equation to answer questions 3 and 4.
MnO2 (s) + Al(s) →
Mn(s) + Al2 O3 (s)
3. If 5.00 moles of Mn(s) were produced, how many moles of Al(s) were required?
a) 3.33 mol
b) 3.75 mol
c) 5.00 mol
d) 6.67 mol
e) none of these
4. If 2.00 moles of MnO2 and 2.00 moles of Al are reacted, how many moles of Al2O3 would be
produced?
a) 1.00 mol
b) 1.33 mol
c) 2.00 mol
d) 3.00 mol
e) 4.00 mol
---------------------------------------------------------------------------------------------------------------5. Consider the reaction below:
4Fe(s) + 3O2 (g) → 2Fe2 O3 (s)
When 50.0 g Fe and 85.0 g of O2 are reacted to produce Fe2O3, is mass conserved in the
reaction? Choose the best answer.
a) No. The different coefficients from reactants to product cause some mass to be lost in
the reaction.
b) No. We started with 135 g of reactants and only produced 71.5 g of product.
c) Yes. The subscripts that appear on the reactant side are also found on the product side,
especially when mole ratios are taken into account.
d) Yes. There are equal masses before and after the reaction, but the mass of leftover
reactant must be taken into account.
e) This cannot be determined because only comparisons can be made in moles when
analyzing balanced chemical reactions, not grams.
CHEMISTRY 101
Hour Exam II
Fall 2006
Page No. 2
6. A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of
solution. What is the molarity of the solution?
a) 1.57 M
b) 1.92 M
c) 3.53 M
d) 333 M
e) none of these
7. Which of the following is the correct net ionic equation for the reaction between barium
hydroxide and sulfuric acid?
a) Ba 2+ (aq) + SO 42- (aq) → BaSO 4 (s)
b) Ba 2+ (aq) + 2OH - (aq) + 2H + (aq) + SO 42- (aq) → 2H 2 O(l) + BaSO 4 (s)
c) 2OH - (aq) + 2H + (aq) → 2H 2 O(l)
d) Ba(OH)2 (aq) + H 2SO 4 (aq) → 2H 2 O(l) + BaSO 4 (s)
e) There is no net ionic equation for this reaction.
---------------------------------------------------------------------------------------------------------------Consider the following four solutions to answer questions 8 – 10.
8. Which of the solutions above contain the greatest number of ions?
a)
b)
c)
d)
e)
Solution #4
Solution #3
Solution #2
Solution #1
At least two of the above solutions contain the greatest number of ions.
9. You pour Solution #1 and Solution #4 together in a large, empty beaker. What is the
concentration of the new HCl mixture?
a) 3.00 M
b) 3.25 M
c) 3.40 M
d) 4.25 M
e) 6.50 M
10. How much water would have to be added to Solution #3 to have the same concentration
of solute as Solution #2?
a) 2.0 L
b) 2.3 L
c) 3.0 L
d) 5.0 L
e) 5.3 L
CHEMISTRY 101
Hour Exam II
Fall 2006
Page No. 3
11. For the unbalanced reaction:
Cr(s) + O 2 (g) → Cr2 O3 (s),
How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium
and excess oxygen gas?
a) 43.8 g
b) 7.50 g
c) 87.7 g
d) 0.144 g
e) 21.9 g
12. Which of the following statements is false for the reaction of hydrogen gas with oxygen
gas to produce water? (a, b, and c represent coefficients)
a H 2 (g) + b O 2 (g) → c H 2 O(g)
a)
b)
c)
d)
The ratio of “a / c” must always equal one.
The sum of a + b + c equals 5 when balanced in standard form.
Coefficient b can equal ½ because coefficients can be fractions.
The number of atoms on the reactant side must equal the number of atoms on the
product side.
e) Subscripts can also be used to balance this equation, just as they can be used to
balance the charges in an ionic compound.
13. Consider the equation: 3A + B → 2C. The molar mass of A is 100.0 g/mol. Which of
the following statements must be 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 less than the molar mass of A, then A must determine how
much C is produced.
e) If the molar mass of B is greater than the molar mass of A, then A must determine
how much C is produced.
14. An air bag is deployed by utilizing the following balanced reaction:
2NaN 3 (s) → 2Na(s) + 3N 2 (g)
What mass of NaN3 must be used to inflate an air bag to 85.0 L at 1.00 atm and 25°C?
a) 3.48 g
b) 27.6 g
c) 151 g
d) 226 g
e) 339 g
CHEMISTRY 101
Hour Exam II
Fall 2006
Page No. 4
15. For the reaction of nitrogen gas with oxygen gas to produce nitrogen monoxide gas,
which of the following “microscopic” reactions correctly represents this chemical
reaction and produces the most product?
a)
b)
c)
d)
e)
CHEMISTRY 101
Hour Exam II
Fall 2006
Page No. 5
16. a. Using different shapes to distinguish between different elements, draw a balanced
equation for the following reaction on the microscopic level. In addition, fill in the
box below to specify what element each shape represents.
NH3 (g) + O 2 (g) →
N 2 (g) + H 2 O(g)
4 points
N=
H=
O=
b. If you react 10.0 g of ammonia with 12.0 g of oxygen gas, how many moles of
nitrogen gas and how many moles of water are produced? Show all of your work
in an organized manner in the space provided.
6 points
Student work will vary here. They need to show the following:
• mass of ammonia and oxygen gas to moles
• determine the limiting reactant as oxygen gas
• show that 0.250 mol N2 is produced
• show that 0.750 mol H2O is produced
c. What is the total volume present (in L) after the reaction is complete assuming the
reaction takes place at 0.°C and 1.00 atm? Show all of your work in an organized
manner in the space provided.
4 points
Student work will vary here. They need to show the following:
• how many moles of NH3 is used up in the reaction (if they didn’t show it
above)
• how many moles of NH3 are leftover
o 0.587 mol NH3 to start – 0.500 mol NH3 used = 0.087 mol NH3
leftover
• use the ideal gas law to solve for the total volume
PV = nRT
nRT (0.250mol + 0.750mol+ 0.087mol)(0.08206)(273)
o V=
=
P
1.00
V = 24.4 L
CHEMISTRY 101
Hour Exam II
Fall 2006
Page No. 6
17. Consider the reaction: 50.0 mL of 2.50 M potassium sulfate is mixed with 100.0 mL of
1.00 M lead(II) nitrate.
3 points
a. Write the balanced molecular, complete ionic, and net ionic equations below. Include
all phases in your reactions.
MOLECULAR:
K SO 4 (aq) + Pb(NO ) (aq) → 2KNO (aq) + PbSO4 (s)
2
3 2
3
COMPLETE IONIC: 2K + (aq) + SO2- (aq) + Pb2+ (aq) + 2NO- (aq) → 2K + (aq) + 2NO- (aq) + PbSO4 (s)
4
NET IONIC:
3
3
22+
SO (aq) + Pb (aq) → PbSO4 (s)
4
b. What mass of solid is produced? Show all of your work in an organized manner in
the space provided.
4 points
9 points
Student work will vary here. They need to show the following:
• moles of K2SO4 or SO2- = 0.125 mol
4
• moles of Pb(NO3)2 or Pb2+ = 0.100 mol
• determine the limiting reactant as Pb(NO3)2 or Pb2+
• show that 0.100 mol PbSO4 is produced
• show that 30.3 g PbSO4 is produced (molar mass of PbSO4 is 303.27
g/mol)
c. What are the concentrations of ions left in solution after the reaction is complete?
Place your final answers in the box below and show all of your work in an
organized manner.
 K +  = 1.67 M
SO 2-4  = 0.167 M
 Pb 2+  = 0.00 M
 NO3-  = 1.33 M
2 mol K +
0.125 mol K 2SO 4 ×
= 0.250 mol K +
1 mol K 2SO 4
[K + ] =
2 mol NO30.100 mol Pb(NO3 ) 2 ×
= 0.200 mol NO31 mol Pb(NO3 ) 2
0.250 mol
0.150 L
[NO3- ] =
0.200 mol
0.150 L
0.125 mol SO 2-4 to start - 0.100 mol SO 2-4 used up = 0.025 mol SO 2-4 leftover
0.025 mol
[SO 2-4 ] =
0.150 L