03/28/16
ANSWER KEY
CHEM 4 Exam 2
Name: __________________________________
Open-End Questions and Problems
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 100 points. (Exam 2 part with multiple choice questions is
worth 50 points. Both parts combined are 150 points total.
1.
(12.5 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)
Chemical Change (a change that involves a
specific pure substance)
Example 1:
ice melts
Example 1:
potassium reacts with water to
produce potassium hydroxide and
hydrogen gas
Example 2:
sulfur burns in air to produce sulfur
dioxide
Example 2:
sugar (sucrose) dissolves in water
Physical Property (specific property of a
specific pure substance)
Chemical Property (specific property of a
specific pure substance)
Example 1:
density of aluminum is 2.70 g/cm3
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
and normal pressure
chemical elements to form any stable
compounds
2.
(12.5 points) Fill the blanks in the following table.
Bromine
Br2
Graphite
C
Hydrogen
sulfide
H2 S
The number of molecules in 1.00 mole of
6.02×1023
none
6.02×1023
The number of atoms in 1.00 mole of
1.20×1024
6.02×1023
1.81×1024
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
160
12.0
34.1
0.0516
0.00545
25.1
The volume in liters (at room temperature
and normal pressure) of 1.00 mole of
3. (12.5 points) Sodium carbonate can neutralize nitric acid. 135 grams of sodium carbonate
is added to a solution that contains 188 g nitric acid. Complete the table below by writing
the balanced chemical equation and calculating the amounts of each reactant and product
before and after the reaction.
Balanced Chemical
Equation
Mass before the
reaction
Molar mass
Moles before the
reaction
Moles consumed
(–) or moles
produced (+)
Moles after the
reaction
Mass after the
reaction
Na2CO3 + 2 HNO3
L.R.
→ 2 NaNO3 +
CO2
+
H2O
135 g
188 g
0 g
0 g
0 g
105.99
g/mol
63.01
g/mol
84.99
g/mol
44.01
g/mol
18.02
g/mol
1.27 mol
2.98 mol
0 mol
0 mol
0 mol
−1.27 mol
−2.55 mol
+2.55 mol
+1.27 mol
+1.27 mol
0 mol
0.43 mol
2.55 mol
1.27 mol
1.27 mol
0 g
27 g
217 g
55.9 g
22.9 g
4. (12.5 points) In the presence of a catalyst, 2.0 moles of carbon dioxide react with 4.0 moles
of hydrogen gas to produce methyl alcohol, CH3OH, and water. Assuming that the reaction
yield was 25%, fill the blanks in the following table.
CO2
Balanced Chemical
Equation
Moles before the reaction
3 H2
L.R.
+
2.0 mol
CH3OH
→
4.0 mol
0 mol
+
H2O
0 mol
Moles consumed (–) or
moles produced (+)
−0.33 mol
−1.0 mol
+0.33 mol
+0.33 mol
Moles after the reaction
1.67 mol
3.0 mol
0.33 mol
0.33 mol
5. (12.5 points) A solution containing 0.200 moles of calcium nitrate is mixed with a solution
containing 0.100 moles sodium phosphate.
(a) Write the full-formula, complete ionic, and net ionic equation (FFE, CIE, and NIE) for
the reaction occurred. In each equation, indicate physical state of each reactant and
product: (s), (l), (g), (aq).
FFE: 3 Ca(NO3)2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaNO3(aq)
CIE: 3Ca2+(aq) + 6NO3−(aq) + 6Na+(aq) + 2PO43−(aq) → Ca3(PO4)2(s) + 6Na+(aq) +6NO3−(aq)
NIE: 3 Ca2+(aq) + 2 PO43−(aq) → Ca3(PO4)2(s)
(b) Complete the following table.
Calcium ion
Moles before the
reaction
0.200 mol
Moles consumed (–)
or moles produced (+)
−0.150 mol
Moles after the
reaction
0.050 mol
Nitrate ion
Sodium ion
spectator ion
spectator ion
0.400 mol
0.300 mol
0 mol
0.400 mol
0 mol
0.300 mol
Phosphate ion
L.R.
0.100 mol
−0.100 mol
0 mol
(c) How many grams of a precipitate are formed? Show work.
0.100 mol PO43− × [1 mol Ca3(PO4)2 / 2 mol PO43−] = 0.050 mol Ca3(PO4)2
0.050 mol Ca3(PO4)2 × (310.17 g/mol) = 15.5 g Ca3(PO4)2
6.
(12.5 points) An experiment is conducted in which varying amounts of solid iron are added
to a fixed volume of liquid bromine. The product of the reaction is a single compound
which can be separated from the reaction mixture and weighed. The graph below shows the
relationship between the mass of iron in each trial versus the mass of the product
compound.
(a) Explain why the graph has a positive slope for low masses of iron and a zero slope
when the mass of iron added becomes larger.
For smaller than 2.0 g of Fe, Fe is the limiting reactant and the bromine is
in excess. The more iron is added to the fixed amount of Br2, the more
product is made. For the masses of Fe greater than 2.0 g, the bromine is
the limiting reactant. Adding 2.0 or more grams of Fe, results in the same
amount of product as the amount of bromine used through the entire
experiment is fixed.
(b) What is the mass in grams of bromine that was used in the experiment? Explain.
When 2.0 g of Fe is used up, 11.0 g of the product is made. The product
can contain only Fe and Br. Therefore, the mass of Br is 11.0 g – 2.0 g =
9.0 g.
(c) What is the empirical formula of the product? Show work.
Fe
2.0 g
×
Br
9.0 g
×
1 mol
55.85
1 mol
79.90 g
= 0.0358 mol
= 0.1126 mol
Formula of the product: FeBr3
0.0358 mol
0.0358
0.1126 mol
0.0358
= 1.0 mol
= 3.1 mol
7. (12.5 points) Briefly answer the following questions.
(a) Why do solid and liquid metals conduct electricity?
Metals have electrons that are free to move between atoms. Those free
electrons are the carriers of the electricity in metals.
(b) Why do liquid (melted) ionic compounds conduct electricity, but solid ionic
compounds do not?
Ionic compounds contain charged particles, ions, in the solid and liquid
states. The ions can freely move and therefore be carriers of electricity in
the liquid, but not in the solid state.
(c) Acids are covalent compounds and are made of molecules, but their aqueous solutions
conduct electricity. Explain.
Molecules of acids react with water molecules in aqueous solutions to
produce ions: hydrogen ions (hydronium ions) and anions.
(d) The color of phenolphthalein, a common acid-base indicator, is pink in an aqueous
solution of ammonia. Explain this fact in words and by writing a chemical equation.
NH3(aq) + HOH(l) ⇌ NH4+(aq) + OH−(aq)
Phenolphthalein turns pink in the presence of OH− ions.
(e) Barium ions are poisonous, but ingestion of barium sulfate causes no harm to the body.
Explain.
Soluble salts of barium are poisonous, because their aqueous solutions will
contain high concentrations of Ba2+. BaSO4 has very low solubility; it’s
ingestion will result in extremely low concentrations in bodily flluids.
8. (12.5 points) A piece of aluminum metal is placed into a solution of hydrochloric acid.
(a) Write the full-formula, complete ionic, and net ionic equation (FFE, CIE, and NIE) for
the reaction occurred. In each equation, indicate physical state of each reactant and
product: (s), (l), (g), (aq).
FFE: 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)
CIE: 2 Al(s) + 6 H+(aq) + 6 Cl− (aq) → 2 Al3+(aq) + 6 Cl− (aq) + 3 H2(g)
NIE: 2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g)
(b) Write the oxidation and reduction half-reactions.
Oxidation
half-reaction: 2 Al(s) → 2 Al3+(aq) + 6 e−
Reduction
half-reaction: 6 H+(aq) + 6 e− → 3 H2(g)