Chem 111 Exam #2
Chapters 4-6
Spring, 2003
Show all work , including any unit conversions for full credit. Use correct number of significant figures.
Potentially useful information: lead can be +2 or +4, tin can be +2 or +4, copper can be +1 or +2 R = 0.0821 L atm/ mole K
CO
3
2 is carbonate, MnO
Specific heats:
4
is permanganate P total
= P
1
+ P
2
+ P
3
+ …
water: 4.184 J/g

C iron: 0.45 J/g

C gold: 0.13 J/g

C
1 Calorie = 1 kcal, 1 cal = 4.184 Joules (J),1lb = 454 g, Q = sm

T

Name_______________________
Chem 111 Exam #2
Chapters 4-6
Spring, 2003
1.
Give formulas for the names or name the following molecules as needed. For iron compound, use both the –ic, -ous system and the roman numeral system. a.
potassium nitrate
KNO
3 b.
chromium (III) carbonate
Cr
2
(CO
3
)
2 c. SiBr
4 silicon tetrabromide d.
FeS iron (II) sulfide and ferrous sulfide e. P
2
O
5 diphosphorous pentoxide f. PbF
4 lead (IV) flouride or plumbic flouride g. Na
3
PO
4 sodium phosphate h.
NH
4
I ammonium iodide i. Cu(OH)
2 copper (II) hydroxide or cupric hydroxide

2. Draw Lewis structures for the molecules below. Give the name of the shape around the central atom in your Lewis structures. a. CO
2 b. CH
2
Cl
2
Cl
H C Cl
O C O linear
H tetrahedral
3. Give oxidation numbers for each atom in the following molecules. a. sodium carbonate b. Fe(MnO
4
)
3
Na +1, O -2, C +4 O -2, Fe +3, Mn +7
4.
Balance the following equations.
10 Na (g) + 2 KNO
3
(s)

1 K
2
O (s) +
1 Ca (s) + 2 H
2
O (l)

1 H
2
(g)
5 Na
2
O (s) + 1 N
2
(g)
+ 1 Ca (OH)
2
(aq)
13 O
2
(g)

8 CO
2
(g) + 10 H
2
O(g) 2 C
4
H
10
(g) +
1 K
2
CO
3
(aq) +
3 O
2
(g) +
2 HCl (aq)

4 Cr (s)
1 CO
2
(g) + 2 H
2
O (l) + 2 KCl (aq)

2 Cr
2
O
3
(s)

5. A mixture of gasses contains 16.0 grams of O
2
, 44.0 g of N
2
, and 44.0 g of H
2
O.
The mixture’s total pressure is 777 torr. What is the partial pressure of each gas? pressure fraction = mole fraction for gas mixture
16.0 g x 1 mole/32.00 g = 0.500 moles
44.0 g x 1 mole/28.01 g = 1.57 moles
44.0 g x 1 mole/18.02 g = 2.44 moles total moles = 4.51 moles pressure of oxygen = 777 torr x 0.500/4.51 = 86.1 torr pressure of nitrogen = 777 torr x 1.57.../ 4.51... = 2.70 x 10 torr pressure of steam = 777 torr x 2.44.../4.51.... = 4.20 x 10 torr
6. Examine the following reaction.
2 AgNO
3
(aq) + 1 BaCl
2
(aq)

2 AgCl (s) + 1 Ba(NO
3
)
2
(aq)
Is it a redox reaction? Why or why not? If you start with 55.9 grams of silver nitrate and
32.5 g of barium chloride and you recover 43.4 g of silver chloride, what is your percent yield? ox numbers for reactants: Ag +1, N +5, O -2, Ba +2, Cl -1 ox numbers for products: Ag +1, N +5, O -2, Ba +2, Cl -1 no change in ox. numbers, so not a redox reaction
55.9 g AgNO3 x 1 mole/169.88 g x 1 BaCl2/ 2 AgNO3 x 208.232 g/mole = 34.3 g BaCl2 so barium chloride is limiting reagent
34.3 g BaCl2 x 1 mole/208.232 g x 2 AgCl/1 BaCl2 x 143.32 g/ mole = 47.16.... g
% yield = actual/ theoretical = 43.4 g/ 47.16.... g = 92.0%

7.
What intermolecular forces are there? List and describe them. How do intermolecular forces affect boiling point and density of substances?
There are three types of intermolecular (between molecules) forces. all three forces make separate molecules cling to each other like socks that have built up a static charge in the dryer:
Hydrogen bonding: the strongest of the three, it is an attraction between a δ+ on a H atom on one molecule and a δ- on another atom (typically N or O) on a different molecule.
Dipolar forces: this is an attraction between between a δ+ on a non hydrogen atom
(carbon, for example) on one molecule and a δ- on another atom (typically N or O) on a different molecule.
Dispersion forces: the weakest of the three forces, this is an attraction between two atoms on different molecules and neither atom has a permanent partial charge.
Since intermolecular forces hold separate molecules together, the stronger the force holding molecules together, the more energy it will take to separate them, so the higher the boiling point the molecule has. Also, if held together tighter (stronger intermolecular forces), the substance is denser.
8. Ammonia is one of the largest products of the chemical industry today. It is used mainly for fertilizer production and is made via what is called the Haber process.
Basically, hydrogen and air are combined in a reaction vessel. The nitrogen in air combines with the hydrogen to make ammonia. Because there is an unlimited amount of air, hydrogen is always the limiting reagent. The balanced equation is as follows:
N
2
(g) + 3 H
2
(g)

2 NH
3
(g)
If 400.0 g of hydrogen gas were consumed in this reaction, how many grams of ammonia
(NH
3
 would be produced? How many grams of nitrogen gas would be consumed



g H2 x 1 mole/2.0158 g x 2 NH3/ 3 H2 x 17.030 g/mole = 2253 g NH3 produced

g H2 x 1 mole/2.0158 g x 1 N2/ 3 H2 x 28.0135 g/mole = 1853 g N2 or, using conservation of matter (mass balance in balanced eqn) mass N2 = mass NH3 – mass H2 = 2253 g- 400.0 g =1853 g N2


A sample of gas has a molecular wt of 32.00 grams/mole. The gas has a volume of
11702 mL under a pressure of 756 torr and a temperature of 4° C. What is the mass of this sample of gas?
756 torr x 1 atm/760. torr = 0.9947... atm
11702 mL x 1 L/1000 mL = 11.702 L mw = g/mole, and PV=nRT, g = mw x mole n= PV/RT (substituting n here into moles for first eqn gives) g = mw x PV/RT =
32.00g/mole x 0.9947...atm x 11.702 L/ ((0.08206 L atm/mole K) x 277K) = 16.4 g
Extra Credit
If 226.4 grams of water at 12.6

C is mixed with 1113 g of iron at 74.9

C, what is the temperature of the water/iron mixture? heat lost by iron = heat gained by water (after changing sign, because heat lost is negative and heat gained is positive)
-Q of iron = Q of water
-smΔT of iron = smΔT of water, ΔT = (Tf –Ti) where Tf is final temperature (final temperature of water and iron are the same when they are mixed together)
-
.45 J/g°C x 1113 g (Tf-74.9°C) = 4.184 J/g°C x 226.4 g (Tf – 12.6)
500.85 J/°C (Tf-74.9°C) = 947.2576 J/°C (Tf – 12.6)
500 J/°C Tf + 38000 °C= 947.3 J/°C Tf – 11900
40000 J = 1450 J/°C Tf
Tf = 28°C