10/28/15
CHEM 60 Exam 2
Open-End Questions and Problems
ANSWER KEY
Please read the following. To receive full credit for a question or a problem, in addition to the
correct answer, you must show a neat, complete, and logical method of solution where each
number is labeled with the appropriate unit and the final answer is rounded to the correct
number of significant digits. The correct answer without any work shown will generally get
zero credit! When an explanation is required, it should be brief, but accurate and complete.
There are 8 questions for a total of 48 points. (Exam 2 part with multiple choice questions is
worth 52 points. Both parts combined are 100 points total.
1. (6 points) Fill the blanks in the following table.
Aluminum
Al
Bromine
Br2
Methane
CH4
none
6.02×1023
6.02×1023
6.02×1023
1.20×1024
3.01×1024
The number of elements in 1.00 mole of
one
one
two
The number of substances in 1.00 mole of
one
one
one
The number of compounds in 1.00 mole of
none
none
one
The mass in grams of 1.00 mole of
27.0
160
16.0
The volume in liters of 1.00 mole of
0.0100
0.0516
24.4
The number of molecules in 1.00 mole of
The number of atoms in 1.00 mole of
2. (6 points) Give the chemical formula for each of the following molecules, ions, or formula
units.
H3PO4
phosphoric acid
H2SO3
sulfurous acid
HNO3
nitric acid
H2PO4−
dihydrogen
phosphate ion
HSO3−
hydrogen sulfite
ion
NO3−
nitrate ion
HPO42−
hydrogen phosphate
ion
SO32−
sulfite ion
NO2−
nitrite ion
PO43−
phosphate ion
S2−
sulfide ion
NH3
ammonia
H2 S
hydrogen sulfide
AlN
aluminum
nitride
ammonium hydrogen
(NH4)2HPO4 phosphate
3. (6 points) Briefly answer the following questions.
(a) Compare the passage of electricity through a wire and through a solution.
The carriers of electricity in a wire are freely moving electrons; the
carriers of electricity in a solution are freely moving ions.
(b) How can it be that aqueous solutions of all ionic compounds conduct electricity but
aqueous solutions of covalent molecular compounds may or may not conduct
electricity?
All ionic compounds are made of ions. When an ionic compound dissolves in
water, the ions are free to move. Most covalent compounds are made of
molecules. Usually, when a covalent molecular compound dissolves in water,
there are only molecules in the solution. In case of some covalent
compounds, acids and ammonia, the molecules react with water to produce
ions. Solutions of those covalent compounds will conduct electricity.
4. (6 points) Give two examples of each. Note: (i) do not to use the same substance as an
example more than once; (ii) paper, plastic, wood, a stick, gasoline, milk, food are not pure
chemical substances and should not be used as examples.
Physical Change (a change that involves a
specific pure substance)
Example 1:
ice melts
Example 2:
sugar (sucrose) dissolves in water
Physical Property (specific property of a
specific pure substance)
Example 1:
density of aluminum is 2.70 g/cm3
Chemical Change (a change that involves a
specific pure substance)
Example 1:
potassium reacts with water to produce
potassium hydroxide and hydrogen gas
Example 2:
sulfur burns in air to produce sulfur
dioxide
Chemical Property (specific property of a
specific pure substance)
Example 1:
chlorine reacts with iron to produce
iron(III) chloride
Example 2:
Example 2:
ammonia is gas at room temperature neon does not react with other
chemical elements to form any stable
and normal pressure
compounds
5. (6 points) Write the full-formula, complete ionic, and net ionic equation (FFE, CIE, and
NIE) for each of the two reactions described below. In each equation, indicate physical
state of each reactant and product: (s), (l), (g), or (aq).
(a) Aqueous solutions of barium nitrate and ammonium phosphate are mixed.
FFE: 3 Ba(NO3)2(aq) + 2 (NH4)3PO4(aq) → Ba3(PO4)2(s) + 6 NH4NO3(aq)
CIE: 3 Ba2+(aq) + 6 NO3−(aq) + 6 NH4+(aq) + 2 PO43−(aq) → Ba3(PO4)2(s) + 6 NH4+(aq) + 6 NO3−(aq)
NIE: 3 Ba2+(aq) +2 PO43−(aq) → Ba3(PO4)2(s)
(b) Aluminum metal is dissolved in sulfuric acid.
FFE: 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
CIE: 2 Al(s) + 6 H+(aq) + 3 SO42−(aq) → 2 Al3+(aq) + 3 SO42−(aq) + 3 H2(g)
NIE: 2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g)
6.
(6 points) A 6.969-gram sample of barium chloride hydrate lost 1.028 grams of water
when heated to constant mass.
(a) What is the formula of the hydrated salt? Show work.
BaCl2
5.941 g ×
H 2O
1.028 g ×
1 mol
208.24 g
1 mol
18.02 g
= 0.02853 mol
= 0.05705 mol
Formula of the hydrate: BaCl2∙2H2O
(b) What is the name of the hydrated salt?
barium chloride dihydrate
0.02853 mol
0.02853
0.05705 mol
0.02853
= 1.000 mol
= 2.000 mol
7. (6 points) Fill the blanks in the following table and answer the additional question below
the table.
Balanced Chemical
Equation
Moles before the reaction
N2(g)
3.0 mol
+
3 H2(g)
→
2 NH3(g)
6.0 mol
0 mol
Moles consumed (–) or
moles produced (+)
− 0.5 mol
− 1.5 mol
+ 1.0 mol
Moles after the reaction
2.5 mol
4.5 mol
1.0 mol
What is the percent yield for the reaction from the above table? Show work.
H2 is the limiting reactant.
% yield = (1.5 mol H2) / (6.0 mol H2) × 100% = 25%
8. (6 points) Use the molecular diagram given below to answer the questions that follow.
Write the balanced chemical equation for the reaction schematically depicted in the
diagram.
2 NH3 + 3 N2O → 4 N2 + 3 H2O
Which substance is the limiting reactant? What is the percent yield of the reaction? Briefly
explain.
N2O is completely used in the reaction. Therefore, N2O is the limiting reactant
and the percent yield is 100%.