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Chem 115 Problem Session #1 Solutions
Chem 112/114 Review and Chem 115 Preparation
All answers must be given with the correct units and number of significant figures. For
questions 1-14, use scrap paper for your work and record only your answers in the spaces
provided. For question 15 through 18, your work must be shown in detail. Questions will be
marked for completeness, however will not be graded. Attach additional pages showing your
work.
1. The chemical formula for lithium sulfite is:
Li2SO3
-
2. The name of the ClO2 ion is:
chlorite
3. The oxidation number of C in H2CO2 is:
+2
4. The number of significant figures in 0.0002470 is:
4
5. When adding the masses 0.056 kg, 145.213 g, and 1635 mg, what is the total
mass in grams given to the correct number of significant figures?
203 g
6. The chemical formula for hydroiodic acid is:
HI(aq)
7. Liquid oxygen boils at 90.2 K. What is its boiling point on the Celsius scale? -183.0 oC
8. The pressure needed to make synthetic diamonds from graphite is 8.0  104 atm.
What is this pressure expressed in Pascals?
8.1 x 109 Pa
9. 1.00 g of, Na2Cr2O7, contains how many chromium atoms?
4.60 x 1021 Cr atoms
10. The mass (in grams) of one molecule of ammonia to two significant figures is: 2.8 x 10-23 g
11. At 25 ºC, bromine is a liquid and has a density of 3.12 g/mL. What volume
is occupied by 100.0g of bromine?
32.1 mL
12. A gas occupies 1.5 L at 1.0 atm at standard temperature. What will the
volume be if both the pressure and temperature are decreased to 80% of its
original pressure and temperature?
1.5 L
13. What is the molarity of a solution that contains 85.0 g of HCl in 275 mL of
solution?
8.48 M
14. Balance the following equations:
a.
P2O5 (s) + 3
H2O (l) →
b.
3 Pb(NO3)2 (aq) + 2
c.
2
Ag+ (aq) +
2 H3PO4 (l)
Na3PO4 (aq) →
Cu (s) → 2
Ag (s) +
Pb3(PO4)2 (s) + 6
Cu2+ (aq)
NaNO3 (aq)
15. Octane (C8H18) is a major component of gasoline. When octane burns in excess oxygen,
water is produced as well as carbon dioxide. (For this reason, clouds of condensed water
droplets are often seen coming from automobile exhausts, especially on cold days.) The
density of gasoline is 0.79 gmL-1.
a. Write the balanced chemical equation for the combustion of octane.
C8H18 + O2  CO2 + H2O basic equation (1 mark for correct combustion)
2C8H18 + 25O2  16CO2 + 18H2O (1 mark for correct balancing)
b. Calculate the mass of water produced from the combustion of 1.0 L of octane.
1.0 L octane…if we assume gasoline is made purely of octane then…
d = m/V
790 g C8H18 x
m = Vd = (1000 mL)(0.79 g/mL) = 790 g C8H18 (1 mark)
1 mol C8 H18 18 mol H 2 O 18.015 g H 2 O
= 1.1 x 103 g H2O
x
x
114.232 g 2 mol C8 H18
1 mol H 2 O
(1 mark for correct stoichiometry and 1 for correct answer)
c. What volume of carbon dioxide (at STP) is produced from the combustion of 1.0 L of
octane?
790 g C8H18 x
PV=nRT
1 mol C8 H18 16 mol CO 2
= 55.326 mol CO2 (1 mark)
x
114.232 g 2 mol C 8 H18
V=
nRT (55.326 mol)(0.082096 Latm/molK)(273.15 K)
x
P
1atm
= 1.2 x 103 L CO2 (1 for using PV=nRT, 1 for correct answer)
16. A 15.00 mL sample of Ca(OH)2 (aq) was titrated with 17.40 mL of 0.234 M HCl (aq).
a. Write the balanced chemical equation for the reaction.
Ca(OH)2(aq) + 2HCl(aq)  Ca2+(aq) + 2Cl-(aq) + 2H2O(l)
b. What is the molarity of the Ca(OH)2 (aq) solution?
n=MV = (0.234 M)(0.01740 L)
= 0.004072 mol HCl x
1 mol Ca(OH) 2
= 0.002036 mol Ca(OH)2
2 mol HCl
M = n/V = (0.002036 mol Ca(OH)2) / 0.01500 L = 0.136 M Ca(OH)2
17. The souring of wine occurs when ethanol, C2H5OH, is converted by oxidation into acetic
acid:
C2H5OH (aq) + O2 (g) → CH3COOH (aq) + H2O (l)
If 100.0 mL of wine (which is 14.2% ethanol by mass) was oxidized by 5.00 g of oxygen,
determine the concentration of the resulting acetic acid assuming complete reaction. Assume
the density of the wine is 1.00 g/mL, and remember to include the water produced during the
oxidation process.
Given : 100.0 mL solution (wine) = 100 g (due to density)
Wine = 14.2 g C2H5OH / 100 g solution (ie. 14.2% by mass)
Therefore: 100 g solution x 14.2 g C2H5OH / 100 g solution = 14.2 g C2H5OH

14.2 g C2H5OH x
1 mol C 2 H 5 OH
1 mol acetic acid(CH 3 COOH)
x
46.069 g C 2 H 5 OH
1 mol C 2 H 5 OH
= 0.3082 mol CH3COOH
1 mol acetic acid(CH 3 COOH)
1 mol O 2
= 0.1563 mol CH3COOH
x
31.998 g O 2
1 mol O 2
Therefore O2 is the limiting reagent…and 0.1563 mol of CH3COOH is produced.

5.00 g O2 x
How much water is formed from the reaction?
1 mol H 2 O
18.015 g H 2 O
= 2.816 g H2O produced
x
1 mol CH 3 COOH
1 mol H 2 O
We assume density of water = 1.00 g/mL so 2.816 mL water produced
0.1563 mol CH3COOH x
Concentration CH3COOH = n/V where V = 100 mL original solution + 2.816 mL water
produced= 102.816 mL = 0.102816 L
0.1563 mol CH3COOH / 0.102816 L solution
= 1.52 M CH3COOH
18. Answer the questions below regarding the following redox reaction.
Mn2+ (aq) + H2O2 (aq)  MnO2 (s) + H2O (l) ( in basic solution)
a. What are the oxidation states of all elements in the redox reaction?
Mn2+
H2O2
Mn : +2

MnO2
H2O
H : +1
O: -2
H: +1
O: -1
Mn: +4
O: -2
b. Which species is the oxidizing agent, and which is the reducing agent?
Mn2+ is oxidized to Mn: +4, therefore it is the reducing agent.
O gets reduced from oxidation state -1 to -2 therefore it is the oxidizing agent.
c. Balance the redox reaction in basic solution using the half-reaction method. Show all
steps to receive full marks.
Oxidation half-rxn
Mn2+ (aq)  MnO2 (s)
Reduction half-rxn
-Balance non-H,O atoms,
Balance O and H
H2O2 (aq)  H2O(l)
2H2O(l) + Mn2+ (aq)  MnO2(s) + 4H+(aq)
2H+(aq) + H2O2 (aq)  2H2O(l)
-Convert H+ to OH- for
Basic solution
Which simplifies to…
4OH- + 2H2O + Mn2+  MnO2 + 4H+ + 4OH2OH- + 2H+ + H2O2 (aq)  2H2O(l) + 2OH4OH- + Mn2+  MnO2 + 2H2O
H2O2 (aq)  2OH-
-Balance electrons
Add together and
Check balance and charges
4OH- + Mn2+  MnO2 + 2H2O + 2e2e- + H2O2 (aq)  2OH2OH-(aq) + Mn2+(aq) + H2O2(aq)  MnO2(s) + 2H2O(l)
Note: Students may also choose to show step 3 at the end…
d. If a typical hydrogen peroxide solution is 5.00% by mass (density of solution = 1.04
g/mL), calculate the volume of Mn(NO3)2 (aq) that may be completely reacted with
250.0 mL H2O2 (aq). The concentration of the Mn(NO3)2 solution is 0.150 M.
Given: Volume of H2O2(aq): 250.0 mL
Concentration of H2O2: 5.00% by mass
Density of solution: 1.04 g/mL
concn Mn(NO3)2(aq): 0.150 M
Find: Volume Mn(NO3)2?
Scheme: find mass H2O2 using given info  moles H2O2  moles Mn2+  moles Mn(NO3)2
Moles Mn(NO3)2  volume Mn(NO3)2
Mass H2O2 solution = V xd = (250.0 mL)(1.04 g/mL) = 260.0 g H2O2 solution
Mass H2O2 = 260.0 g H2O2 solution x
Moles H2O2 = 13.00 g H2O2 x
5.00 g H 2 O 2
= 13.00 g H2O2
100 g solution
1 mol H 2 O 2
= 0.3821 mol H2O2
34.016 g H 2 O 2
From the balanced chemical equation, it is seen that the ratio of H2O to Mn2+ is 1:1
(Must show this in their response, either in wording or in calculation)
Therefore, 0.3821 mol H2O2 x
1 mol Mn(NO3 ) 2
1 mol M n2
x
= 0.3821 mol Mn(NO3)2
1 mol H 2 O 2
1 mol Mn2
M = n/V
and V of Mn(NO3)2 =
n ( Mn(NO3 ) 2 )
0.3821 mol Mn(NO3 ) 2
=
= 2.547 L of Mn(NO3)2
M ( Mn(NO3 ) 2 )
0.150 M
= 2.55 L Mn(NO3)2