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Chem 104 Exam #3
Chapters 5-6
Show all work, including any unit conversions for full credit. Use correct number of sig figs.
Soluble
K+, Li+, Na+, NH4+
NO3-, CH3COOHalides (Halogens)
Sulfates
Exceptions:
none
none
Ag, Pb, Hg
Ba, Sr, Pb, Hg
Insoluble
Carbonates
Phosphates
Exceptions:
2-
(CO3 )
(PO43-)
Group 1A and NH4+
Group 1A and NH4+
R= 0.08206 L atm/(K moles), specific heat water = 4.184 J/g°C, 1 cal = 4.184 J
7.60 x 102 torr =7.60 x 102 mm Hg = 1.00 atm
heat fusion (water) = 8.0 x 101 cal/g, heat vaporization (water) = 540 cal/g
Oxidation number rules
1. anything neutral and by itself, ox # = 0
2. any simple ion, the ox # = charge on the ion
3. in a molecule, H is always +1, O is always -2
4. the total of all oxidation numbers in a molecule/polyatomic ion = net charge
Name_______________________
Chem 104 Exam #3
Chapters 5-6
1. For each of the reactions below which produces a precipitate, write the formula of the solid
precipitate that is formed. If no precipitate is formed, write NR. Then chose one that gave a
precipitate and write balanced molecular, total ionic, and net ionic equations for it. Include
physical states for all chemicals.
a. silver (I) nitrate (aq) + potasium carbonate (aq)
2 AgNO3 (aq) + K2CO3 (aq) → Ag2CO3 (s) + 2 KNO3 (aq)
2 Ag+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + CO32- (aq) → Ag2CO3 (s) + 2 K+ (aq) + 2 NO3- (aq)
2 Ag+ (aq) + CO32- (aq) → Ag2CO3 (s)
b. AgNO3 (aq)
+
NaCH3COO (aq)
c. BaCl2 (aq)
+
MgSO4 (aq)
NR
BaCl2 (aq)
+
MgSO4 (aq) → BaSO4 (s) + MgCl2 (aq)
Ba2+ (aq) + 2 Cl- (aq) + Mg2+ (aq) + SO42- (aq) → BaSO4 (s) + Mg2+ (aq) + 2 Cl- (aq)
Ba2+ (aq) + SO42- (aq) → BaSO4 (s)
2. A sample of O2 gas has a volume of 11401 mL, a pressure of 2.72 atm, and has a temperature
of 25°C. What is the mass of the gas?
mw = mass/mole; mass = mw x n
PV=nRT; n = PV/(RT)
mass = mw x PV/(RT) = 32.00 g/mole x 2.72 atm x 11.401 L / (0.08206 Latm/moleK x 298 K =
= 40.6 g of oxygen
3. Using the balanced equation below, how many grams of water are made when 136.8 g of
magnesium hydroxide are reacted with 167.4 g of H3PO4?
2 H3PO4
+
3 Mg(OH)2

6 H2O +
1 Mg3(PO4)2
Mg(OH)2 = (24.305 + 2 x 15.999 + 2 x 1.0079) = 58.319 g/mole
H3PO4 = ( 3x 1.0079 + 30.974 + 4 x 15.999) = 97.994 g/mole
H2O = 2 x 1.0079 + 15.999 = 18.015 g/mole
136.8 g x 1 mole/ 58.319 g x 6 H2O/ 3 Mg(OH)2 x 18.015 g/mole = 84.52 g
167.4 g x 1 mole/97.994 g x 6 H2O/2 H3PO4 x 18.015 g/mole = 92.32 g
So magnesium hydroxide is the limiting reagent and 84.52 g of water are produced
4. Explain why methyl ether (H3C-O-CH3) and ethanol (CH3-CH2-OH) have such vastly
different boiling points (-25º C vs 78º C) even though they have the exact same chemical
formula (C2H6O).
Since both molecules have the same molecular mass, they have roughly the same London
dispersion forces holding them together
Methyl ether has polar bonds (2 C-O), but does not have a partial plus on hydrogen, only carbon.
As a result, its strongest intermolecular force is dipole-dipole attractions. It has a lot of
hydrogens, but cannot hydrogen bond because there aren’t any partial pluses on those H’s.
Dipole dipole interactions are the second stronges intermolecular force.
Ethanol has just as many polar bonds (C-O and O-H), but it has a partial plus on hydrogen. That
means that this molecule can hydrogen bond to other ethanol molecules. H-bonding is the
strongest intermolecular force.
Because the ethanol molecules can H-bond to hold themselves together and the methyl ether
molecules can only dipole-dipole to hold on to each other, it takes less energy to separate the
methyl ether molecules from each other and they have a lower bp
5. The tarnish on silverware is usually because of a silver sulfide (Ag2S) layer on the surface. It
can be removed with some aluminum foil in a solution with the silverware. The reaction is:
3 Ag2S (s)
ON
+1 -2
+
2 Al (s)
0
→
6 Ag (s)
+
0
1 Al2S3 (s)
+3 -2
First, balance the equation. Second, give oxidation numbers for all atoms in this reaction
(these sulfur compounds are ionic). Third, tell me what is oxidized, what is reduced, what is the
oxidizing agent, and what is the reducing agent.
Ag in Ag2S went from + 1 to 0; it was reduced, so it was the oxidizing agent
Al went from ON of 0 to +3, so it was oxidized and was the reducing agent
6. A steel tank of nitrous oxide (N2O) gas is filled with 50.0 moles to 44.0 atm at 25ºC in the
dentist’s office. The N2O is slowly used up such that the temperature never changes. When
the pressure drops to 17.76 atm, how many moles are left in the tank?
(PV/nT)1 = (PV/nT)2; since V and T don’t change, we have (P/n)1 = (P/n)2;
n2 = P2 x n1/P1 = 17.76 atm x 50.0 moles/44.0 atm = 20.2 moles left in tank
7. A gas container is filled with inert gasses. The partial pressure of Argon is 1533 torr, Neon
has a partial pressure of 236 mm Hg, while Krypton has a partial pressure of 1.666 atm. If
these are the only gasses in the tank, and there are 8.78 moles of gas in the tank, how many
moles of each of the three gasses are present?
Ar →
1533 torr
Ne →
236 torr
Kr → 1.666 atm x 760 torr/atm =
1266 torr
3035 torr total
Dalton’s law of partial pressures sez that pressure fraction = mole fraction so…
Ar = (1533 torr/3035 torr) x 8.78 moles = 4.43 moles
Ne = (236 torr/3035 torr) x 8.78 moles = 0.683 moles
Kr = (1533 torr/3035 torr) x 8.78 moles = 3.66 moles
Extra Credit:
51.5 g of H2O goes from ice at a temperature of -13°C to steam at a temperature of
102°C. This happens at 1.00 atm. How much heat (in Joules) was absorbed by the H2O?
oops! forgot to include specific heat of ice and steam on potentially useful information (both at
0.48 cal/g ºC)
heat from -13ºC to 0ºC: Q = smΔT = 0.48 cal/gºC x 51.5 g x 13 ºC = 321.36 or 320 cal
heat to melt:
Q = Hf x m = 80 cal/g x 51.5 g = 4120 or 4100 cal
heat from 0 to 100ºC:
Q = smΔT = 1.00 cal/gºC x 51.5 g x 100 ºC = 5150 cal
heat to vaporize:
Q = Hv x m = 540 cal/g x 51.5 g = 27810 or 27800 cal
heat from 100 to 102ºC:
Q = smΔT = 0.48 cal/gºC x 51.5 g x 2 ºC = 49.44 or 50 cal
total = 5.30 x 104 cal
5.30 x 104 cal x 4.184 J = 2.22 x 105 J