Chm 101 – 1,2
Exam 2 100 Points
10-14-15
Name
Multiple Choice: circle the one best answer (4 pts. each)
1) The number of carbon dioxide molecules present in 3.00 moles of CO2 is:
a) 1.81 x 1024
b) 3.00
c) 6.02 x 1023
d) 1.20 x 1046
e) none of the above
2) What is the number of oxygen atoms that constitute a mass of 32.00 g?
a) 2
b) 8
d) 6.02 x 1023
c) 16
e) none of the above
3) What is the term that expresses the simplest whole number ratio of atoms in a molecule?
a) chemical formula
b) empirical formula
c) molecular formula
d) mass ratio composition
e) percent composition
4) The empirical formula for a compound containing only nitrogen and oxygen is NO2. If the molar mass
of the compound is 138.03 g/mol, the molecular formula is:
a) NO2
b) N2O4
c) N2O6
d) N2O
e) none of the above
5) Which of the following represents the chemical formula for chromium(III) sulfate?
a) Cr2(SO4)3
b) Cr3(SO4)2
c) CrSO4
d) Cr3SO4
6) Which of the following ionic compounds is insoluble in water?
a) Na3PO4
b) AgNO3
c) NaCl
d) CaCO3
e) MgCl2
7) Which of the following will have the greatest mass?
a) 2 mol H2O
b) 2 mol NaCl
c) 2 mol O2
d) all have the same mass
8) The molar mass of iron(III) nitrate is
a) 178.87 g/mol
b) 115.88 g/mol
c) 241.86 g/mol
d) none of the above
9) The molarity of a solution prepared by dissolving 0.145 mols Na2SO4 in enough water to make 500 mL
of solution is:
a) 0.145 M
b) 2.90 x 10-4 M
c) 0.290 M
d) 0.193 M
10) How many total moles of ions are released when 0.25 mol Na3PO4 (s) dissolves completely in water?
a) 0.25 mols
b) 0.50 mols
c) 0.75 mols
d) 1.00 mols
11) The reducing agent in the following electrochemical reaction is:
6 Fe2+ + Cr2O72- + 14 H+  6 Fe3+ + 2 Cr3+ + 7 H2O
a) O2-
b) Cr6+
c) H+
d) Fe2+
e) Cr3+
12) The following reaction is best described as:
H2SO4 (aq) + 2 KOH (aq)  K2SO4 (aq) + 2 H2O (l)
a) oxidation-reduction
b) precipitation
c) combustion
d) acid-base neutralization
For the following numerical problems, you must show all of your work to receive full credit. Watch
significant figures!
13) Ethanol, C2H6O, is the “alcohol” in alcoholic beverages. Calculate the number of grams of carbon in
75.3 g of ethanol. (10 pts.)
75.3 g C2H6O
1 mol C2H6O
2 mol C
12.01 g C
46.08 g C2H6O
1 mol C2H6O
1 mol C
= 39.3 g C
14) The reaction between the weak acid, citric acid, and the weak base, sodium bicarbonate, is shown in
the (unbalanced) equation below
3 NaHCO3 (aq) + H3C6H5O7 (aq)  3 CO2 (g) + 3 H2O (l) + Na3C6H5O7 (aq)
If 1.00 g of citric acid and 3.00 g of sodium bicarbonate are allowed to react:
a) Identify the limiting reactant, if any. (6 pts.)
Count how many mols of each reactant.
1.00 g CA
3.00 g SB
1 mol CA
192.14 g CA
1 mol SB
84.01 g SB
-3
= 5.20 x 10 mols CA
-2
= 3.57 x 10 mols SB
-2
3.57 x 10 mols SB
-3
5.20 x 10 mols CA
= 6.86
There is more than 3x as much SB as CA
so the CA runs out first. CA is the limiting
reactant.
b) Calculate the number of grams of H2O (l) that can be produced. (6 pts.)
The amount of product formed is always based on the limiting reactant!
1.00 g CA
1 mol CA
192.14 g CA
3 mol H2O
18.02 g H2O
1 mol CA
1 mol H2O
= 0.281 g H2O
15) The active ingredient in many hair bleaches is hydrogen peroxide. The amount of H2O2 in 14.8 g of
hair bleach was determined by titration with standard potassium permanganate solution:
2 KMnO4 (aq) + 5 H2O2 (aq) + 6 H+(aq)  5 O2 (g) + 2 Mn2+(aq) + 2 K+(aq) + 8 H2O(l)
If 43.20 mL of 0.105 M KMnO4 was required to reach the endpoint, calculate the mass percent of
hydrogen peroxide in the sample. (10 pts.)
43.20 mL KMnO4
5 mol H2O2
0.105 mol KMnO4
103 mL
0.3859 g H2O2
2 mol KMnO4
34.02 g H2O2
1 mol H2O2
= 0.3859 g H2O2
X 100 = 2.61 % H2O2
14.8 g bleach
16) Complete and balance the following reactions and write the net ionic equation for each. (10 pts.)
2 NaClO4 (aq) + CaCl2 (aq)  2 NaCl (aq) + Ca(ClO4)2 (aq)
All reactants and products are soluble in water. No net reaction takes place.
KOH (aq) +
Base
OH-(aq) +
CH3COOH (aq)  Na CH3COO (aq) + H2O (l)
Acid
H+(aq)  H2O (l)
Salt
Water
17) A stock solution of ammonia has a concentration of 14.8 M. How many milliliters of this solution
should you dilute to make 1000.0 mL of 0.250 M NH3 (aq). (6 pts.)
M1V1 = M2V2
V1 =
M2V2
M1
0.250 M x 1000.0 mL
=
= 16.9 mL
14.8 M
Extra Credit: (Choose one, 10 pts. all or none)
1) Some sulfuric acid is spilled on a lab bench. It can be neutralized by sprinkling sodium bicarbonate on it
and mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid as follows
2 NaHCO3 (s) + H2SO4 (aq)  Na2SO4 (aq) + 2 H2O (l) + 2 CO2 (g)
Sodium bicarbonate is added until the fizzing due to formation of CO2 (g) stops. If 27 mL of
6.0 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to
neutralize the acid?
Keep working on this !
2) A person suffering from hyponatremia has a sodium ion concentration in the blood of 0.118 M and a
total blood volume of 4.6 L. What mass of sodium chloride would need to be added to the blood to bring
the sodium ion concentration up to 0.138 M, assuming no change in blood volume?
Keep working on this !