AP Chemistry
Unit 3- Homework Problems
Gas Laws and Stoichiometry
STP
1. What is standard pressure for each of the following: atm, mm Hg, Torr, kPa, PSI
1 atm = 760 mm Hg = 760 torr = 101 kPa = 14.7 psi
2. Convert each of the following:
a. 700 mm Hg to atm 0.92 atm
b. 1.5 atm to Torr
1140 Torr
c. 97 kPa to PSI
14.1 psi
d. 750 Torr to mm Hg
750 mm Hg
3. What is a barometer and how does it work? Measures atmospheric pressure. Air pressure
presses down on a pool of mercury that is pushed up an evacuated
4. What pressure is on this open-ended manometer? 810 mm Hg (50 + 760)
50 mm
5. What would the pressure be if #4 was a closed-end vacuum manometer? 50 mm Hg
6. What is standard temperature? 0 oC or 273 K
7. How is the Kelvin scale different from the Celcius and Fahrenheit scale? There are no
negative numbers. Zero on the Kelvin scale is the lowest you can go.
8. Why is the Kelvin scale called the “absolute” scale? Zero on the Kelvin scale is as low as you
can go. Zero means zero.
Pressure and the Atmosphere
1. What is the relationship between altitude and atmospheric pressure? As altitude increases,
pressure decreases.
2. Why does it take longer to cook food in Denver than in Palatine? There is less atmospheric
pressure in Denver due to the high altitude. Thus, it takes less temperature to get water’s
vapor pressure to equal atmospheric pressure. Thus, water boils at a lower temperature and
thus food takes longer to cook.
3. How does a pressure cooker work? Artificially increases atmospheric pressure above 1 atm.
Thus water must get hotter for VP to equal AP. Thus water boils at a higher temperature and
thus food cooks faster.
4. How far under water do you have to go to increase pressure by 1 atm? About 30 feet
5. What are “The Bends”? At very low depths underwater pressure is extremely high. The N2
you normally breathe in air usually goes in and out of the lungs without being absorbed by the
bloodstream. At higher pressures, due to Henry’s Law, the solubility of the gases increases
and thus your bloodstream has N2 gas in it. When you surface, the pressure decreases and the
solubility decreases. Thus you have N2 (g) coming out of your bloodstream. Because your
blood goes everywhere in your body, N2 (g) is bubbling out of your brain, heart, organs, and
anywhere else blood goes. This causes massive problems and could kill you.
Gas Laws
1. A gas is at 10 L and 700 mm Hg. What is the volume at 350 mm Hg? 20 L
2. A gas is at 80 mL and 75 kPa. What is the pressure at 50 mL? 120 kPa
3. A gas is at 2.4 L and 25 oC. What volume is the gas at 50 oC? 2.60 L
4. A gas is at 400 mL and 15 oC. What temperature (oC) will cause the gas to be 100 mL? 72 K
or -201 oC
5. A gas is at 2.3 atm and 0oC. What pressure will cause the gas to be -50 oC? 1.88 atm
6. A gas is at 18 psi and 20 oC. What temperature will cause the gas to be 50 psi? 814 K or 541
o
C
7. A gas has 2.4 moles and 5 L. How many moles will cause the gas to be 8 L? 3.4 moles
8. A gas has 1.7 moles and 350 mL. How many mL will the gas occupy if it is 7.8 moles
instead? 1606 mL
9. A gas has 5.4 L at 800 mm Hg and 20 oC. What volume will the gas occupy if the pressure
drops to 600 mm Hg while the temperature increases to 80 oC? 8.67 L
10. A gas has 650 mL at 15 oC and 3.4 atm. What pressure will the gas be if the volume
increases to 1000 mL while the temperature drops to -20 oC? 1.94 atm
11. A gas has 400 mL at 5 oC and 6.5 psi. What temperature will the gas be if the volume is
dropped to 200 mL while the pressure is increased to 20 psi? 428 K or 155 oC
12. For each of the questions above, state whether the gas law is: Charles’, Gay-Lussac’s,
Boyle’s, Avagadro’s, or Combined.
1. Boyles
2. Boyles
3. Charles
4. Charles
5. Gay-Lussac
6. Gay-Lussac
7. Avagadro’s
8. Avagadro’s
9. Combined
10. Combined
11. Combined
13. Write the Ideal Gas Law. What is the value for R? What are the units of R?
PV = nRT R = 0.0821 Latm/molK
14. What is the pressure on an ideal gas if 0.25 moles of it occupies 3 L at 300 K? 2.05 atm
15. What volume does 50 grams of CO2 occupy if it is at 0oC and 700 mm Hg? 27.65 L
16. How many grams of CH4 are present if 300 mL of it is at 120 kPa and -25 oC? 0.28 grams
17. What is the density of H2 gas at STP? 0.089 g/L
18. What is the density of air (assume 80% N2 and 20% O2) at STP? 1.285 g/L
19. What is the density of Xe gas at 22 oC and 740 mm Hg? 5.27 g/L
20. What is the molecular weight of a gas if 2.3 grams of it occupy 230 mL at a pressure of 750
mm Hg and a temperature of 75 oC? 290 g/mole
21. A gas is 11.8% C, 69.6% Cl and 18.6% F. If 0.107 grams of it fills a 458 mL flask at 25 oC
at a pressure of 21.3 mm Hg, what is the molecular formula of the compound?
Empirical CCl2F = 102 g/mole
MW = 204 so molecular = C2Cl4F2
22. A gas is 25.2% S and 74.8% F. If 0.0955 g are put in an 89 mL flask at 45 oC and a pressure
of 83.8 mm Hg, what is the molecular formula of the gas?
Empirical SF5 = 127.1 g/mole
MW = 254 so molecular = S2F10
23. State Dalton’s Law of Partial Pressures. Ptotal=  Pindividual
24. A container has 0.5 atm of H2, 2.5 atm of He, and 3 atm of O2. What is the Ptotal in the
container? Ptotal = 0.5 + 2.5 + 3 = 6 atm
25. A container has 10 grams of N2, 100 grams of Xe, and 42 grams of Ar. What is the Ptot in
the container if the volume is 4 L and the temperature is 50 oC?
nN2 = 0.357 moles nXe = 0.762 moles
nAr = 1.053 moles
ntotal = 2.17 moles
Ptotal = 14.4 atm
26. A container has 20 grams of O2, 40 grams of CH4, and 10 grams of He at a Ptot of 600 mm
Hg. What is the partial pressure of each gas?
nO2 = 0.625 moles nCH4 = 2.5 moles
nHe = 2.5 moles
ntotal = 5.625 moles
PO2 = (0.625/5.625)* 600 mm Hg = 66.7 mm Hg
PCH4 = (2.5/5.625)* 600 mm Hg = 266.7 mm Hg
PHe = (2.5/5.625)* 600 mm Hg = 266.7 mm Hg
27. Which of the following gases would be nearly impossible to collect over water? Why?
N2, HCl, NH3, Cl2 HCl and NH3 would be nearly impossible to collect over water because
they are soluble in water. They will dissolve in it to make a solution so wouldn’t come out
the other end.
28. Hydrogen gas is collected over water at 15 oC. If the gas was collected at 800 mm Hg
atmospheric pressure, what pressure of H2 gas was collected?
VPH2O at 15 oC = 12.8 mm Hg so PH2 = 800 -12.8 = 787.2 mm Hg
29. Oxygen gas is collected over water at 22 oC. If the gas was collected in a 250 mL container
at 740 mm Hg atmospheric pressure, how many grams of oxygen were collected?
VPH2O at 19.8 mm Hg so PO2 = 740 – 19.8 = 720.2 mm Hg
0.313 grams of O2
Kinetic Molecular Theory
1. What are the four ideas of the kinetic molecular theory?
a. Gases consist of particles that are in random, rapid motion
b. Kinetic energy of the gas is directly proportional to temperature
c. Gases collide with each other and their container with no loss of energy (elastic)
d. Gas particles occupy no volume relative to the overall volume of the container itself
2. Which two of the ideas in #1 are not technically true?
c&d
3. What is the Van der Waal’s equation? What does it correct for?
[P+ a(n/V)2][V-bn] = nRT It corrects for c & d above
4. Which gas in each pair would deviate most from ideality and why?
a. N2 vs. CO CO because it is polar at the same mass
b. CH3OH vs. O2 CH3OH has H-bonds at the same mass
c. Ne vs. Xe They are both non-polar but Xe is bigger so has more London forces
5. Two equal balloons are made; one with Helium and one with Neon. How much faster does
the Helium balloon deflate compared to the Neon? 2.25 times as much
6. SO2 diffuses 1.5 times as fast as an unknown gas. What is the molar weight of the unknown
gas? 144 g/mole
7. If BH3 is put into one end of a tube marked 0 cm and NF3 is put into the other end marked
100 cm, at what distance mark on the tube do they meet? 69.4 cm
8. What is the RMS speed of O2 gas at STP? 461 m/s
9. What is the RMS speed of Rn gas at STP? 175 m/s
10. You have four gas samples:
a. 1 L H2 at STP
b. 1 L of Ar at STP
c. 1 L of H2 at 27 oC and 760 mm Hg
d. 1 L of He at 0 oC and 900 mm Hg
i) Which sample has the most atoms or molecules? Sample D
ii) Which sample has the least atoms or molecules? Sample C
iii) Which sample has the largest mass? Sample B
iv) Which sample has the largest KE? Sample C
11. Five identical balloons are each filled to the same volume at the same temperature and
pressure but are filled with CO2, O2, He, N2, and CH4.
a. Which has the greatest mass? CO2
b. Compare the KEavg of the molecules in each of the balloons. Explain. They are all the
same as they are at the same T
c. Which should deviate most from ideality? Why? All non-polar so CO2 has largest mass
d. All of the balloons decrease in size over time. Which balloon will be the smallest? Why?
He will effuse through the balloon fastest as it is the smallest
12. CO and CO2 are put in separate containers of the same size and the same temperature. The
pressure of CO gas is 2 atm while the pressure of the CO2 gas is 1 atm.
a. Compare the KEavg of the 2 samples. Explain. KE is the same as both are at same T
b. Which gas has a greater RMS speed? Explain why. CO has smaller mass so moves faster
c. Which container contains a greater number of molecules. Justify your answer. CO is at
same conditions but has 2 atm while CO2 has only 1 atm. The extra pressure must be
coming from extra molecules
Molarity
1. What is the definition of M? How do you make a M solution?
M = moles solute/L solution
To make a M solution you figure out how much solute you have to get into the solution and
dissolve it in a little of the solvent. Once the solute is all dissolved, you add additional solvent
until the total volume of the solution is what you require it to be. Note the formula tells you
how much solute is present, how much solution is present, but NOT how much solvent is
needed.
2. What is the M of a solution that has 2.5 moles of C2H5OH dissolved in 2500 mL of solution?
1M
3. What is the M of a solution that has 100 grams of Na2CO3 dissolved in 475 mL of solution?
1.99 M
4. How many grams of K3PO4 must be dissolved in 250 mL total to make a 0.75 M solution?
39.8 g
5. How many mL are needed to dissolve 35 grams of CaCl2 into a 0.025 M solution? 12,600
mL
6. What is the concentration of each of the ions in each of the following solutions:
a. 0.50 M KI
[K+1] = 0.50 M
[I-1] = 0.50 M
b. 1.25 M Li2SO4 [Li+1] = 2.50 M
[SO4-2] = 1.25 M
c. 0.68 M Na3PO4 [Na+1] = 2.04 M
[PO4-3] = 0.68 M
d. 0.25 M Al2(SO4)3
[Al+3] = 0.50 M
[SO4-2] = 0.75 M
7. If 20 mL of 6 M HCl is diluted to a total volume of 100 mL, what is the new concentration?
1.2 M
8. If 127 mL of 0.125 M H2SO4 needs to be diluted to 0.057 M, what is the new volume? 279
mL
9. If 100 mL of 2.5 M solution A is mixed with 100 mL of 1.6 M solution B, what is the new
concentration of each solution?
Solution A = (2.5 M)(100/200) = 1.25 M
Solution B = (1.6 M)(100/200) = 0.80 M
10. If 57.2 mL of 0.125 M solution A is mixed with 32.9 mL of 0.085 M solution B, what is the
new concentration of each solution?
Solution A = (0.125 M)(57.2/90.1) = 0.0794 M
Solution B = (0.085 M)(32.9/90.1) = 0.0310 M
Saturation Curves
1. What is the solubility of NH4Cl at 40 oC? @50 g NH4Cl/100 g H2O
2. How many grams of KCl will dissolve in 300 g H2O at 80 oC?
solubility @55 g KCl/100 g H2O so 165 g KCl in 300 g H2O
3. A solution has 50 grams of LiCl dissolved in 100 g H2O at 50 oC. Is it saturated, unsaturated,
or supersaturated? Solubility is @95 g/100 g H2O so unsaturated
4. 500 grams of H2O were made into a saturated solution of NaNO3 at 80oC. The solution was
then cooled to 20 oC. How many grams of NaNO3 ppt out of solution? 150 g/100 g H2O at
80oC to 90 g/100 g H2O at 20 oC so difference of 60 g/100 g H2O so 300 g in 500 g H2O ppt
out
5. On the graph above, draw in a line which represents the solubility of gases such as CO2. The
blue line above represents gases as the higher the temperature, the lower the solubility of
gases.
Colligative Properties
1. List the following solutions in order of increasing melting point:
a. 0.1 m sugar
b. 0.1 m NaCl
c. 0.08 m CaCl2
d. 0.04 m Na2CO3
0.08 m CaCl2 < 0.1 m NaCl < 0.04 m Na2CO3 < 0.1 m sugar
2. List the following solutions in order of decreasing freezing point:
a. 0.20 m C2H4(OH)2
b. 0.12 m K2SO4
c. 0.10 m MgCl2
d. 0.12 m KBr
0.20 m > 0.12 m KBr > 0.10 m MgCl2 > 0.12 m K2SO4
3. What happens to the vapor pressure of a solvent when a solute is dissolved in it?
Vapor pressure decreases as solute concentration increases
Combination Problems
1. Two containers each have a volume of 1.0 L at 298 K. Container A has 0.10 moles of N2 (g)
and container B has 0.10 moles of H2 (g). The average kinetic energy of the N2 (g) molecules
is 6.2 x10-21 J. Assume both gases exhibit ideal behavior.
a. Is the pressure in the container of H2 greater than, less than, or equal to the pressure of the
N2? Justify your answer. Pressure should be the same as all conditions are equal
b. What is the average kinetic energy of the H2 (g) molecules. Justify your answer.
6.2x10-21 J because same temperature means it’s the same KE
c. The molecules of which gas, N2 or H2have the greater average speed? Justify. H2 has
greater speed as they are at the same KE so H2 is less massive so moves faster
d. What change could be made that would decrease the average kinetic energy of the N2
molecules? Justify. Only way to decrease the KEavg is to lower the Temperature.
e. If the volume of the container of H2 was decreased to 0.50 L at 298 K, what effect would
that have on:
i) The pressure in the container. Justify. Pressure would double by Boyle’s Law
ii) The average speed of the H2 molecules. Justify. Speed of the molecules wouldn’t
change as temperature hasn’t changed
2. A rigid 5.0 L cylinder contains 24.5 g of N2 and 28.0 g of O2.
a. Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K.
24.5 g N2 (1 mole/28 g) = 0.875 moles N2
28 g O2 (1 mole/32 g) = 0.875 moles O2
Ptot = nRT/V = (0.875 + 0.875)(0.0821)(298 K)/5 L = 8.56 atm total
b. The temperature of the gas mixture in the cylinder is decreased to 280 K. Calculate:
i) The mole fraction of N2 in the cylinder
XN2= (0.875)/(0.875+0.875) = 0.50
ii) The partial pressure, in atm, of N2 in the cylinder
4.28 atm N2 at 298 K so
(4.28 atm/298 K) = (x atm)/280 K
4.165 atm
c. If the cylinder develops a pinhole-sized leak and the gases begin to escape, would the ratio
of moles of N2/moles of O2 in the cylinder increase, decrease, or remain the same. Justify
your answer.
Ratio should decrease as moles N2 will be less as it is moving faster through the container
than O2
3. Concentrated sulfuric acid (18.4 M H2SO4) has a density of 1.84 g/mL. After dilution with
water to 5.2 M, the solution has a density of 1.38 g/mL.
a. Calculate the volume of concentrated acid needed to make 1.0 L of 5.2 M H2SO4
(18.4 M)(X L) = (5.2 M)(1 L)
0.283 L of 5.2 M needed
b. Determine the mass % of H2SO4 in the original concentrated solution.
1805 g H2SO4/1840 g solution *100% = 98.1%
c. Calculate the volume of 5.2 M H2SO4 that can be completely neutralized with 10.5 g of
NaHCO3.
H2SO4 + 2 NaHCO3  2 H2CO3 + Na2SO4
10.5 g NaHCO3 *(1mole/84 g) = 0.125 moles NaHCO3
0.125 moles NaHCO3 * (1 mole H2SO4/2 NaHCO3) = 0.0625 moles H2SO4
0.0625 moles (1 L/5.2 moles) = 0.0120 L or 120 mL
d. What is the molality of the 5.2 M H2SO4 solution?
5.2 moles/0.8699 kg H2O = 5.978 m
Types of Reactions & Balancing
1. For each of the equations below, decide whether the reaction is an example of: synthesis,
decomposition, single replacement, double replacement, or combustion
SR a. CaI2 + Br2  CaBr2 + I2
Comb. CH4 + 2 O2  CO2 + 2 H2O
Syn c. 2 Al + 3 Cl2  2 AlCl3
Syn d. 6 K + N2  2 K3N
DR e. 3 MgCl2 + 2 K3PO4  Mg3(PO4)2 + 6 KCl
Dec f. CaCO3  CaO + CO2
Dec g. NH4OH  NH3 + H2O
Syn h. 2 Li + F2  2 LiF
Comi. CS2 + 3 O2  CO2 + 2 SO2
SR j. Zn + Cu(NO3)2  Zn(NO3)2 + Cu
2. Balance the following equations:
a. Fe2O3 + 3 Mg  3 MgO + 2 Fe
b. SF4 + 2 H2O  SO2 + 4 HF
c. BF3 + 3 H2O  3 HF + H3BO3
d. C6H5CH3 + 9 O2  4 H2O + 7 CO2
e. 3 Fe + 4 H2O  Fe3O4 + 4 H2
3. Predict the products and balance the equation:
a. Si + 2 Cl2  SiCl4
b. Na2SO4 + BaCl2  BaSO4 + 2 NaCl
c. 2 CaO  2 Ca + O2
d. Zn + 2 AgNO3  Zn(NO3)2 + 2 Ag
e. 2 C8H18 + 25 O2  16 CO2 + 18 H2O
Stoichiometry
1. For the reaction:
4 Al + 3 O2  2 Al2O3
a.
b.
c.
d.
How many moles of Al are needed to react with 9 moles of O2? 12
How many moles of Al2O3 can be made from 3.2 moles of O2 and excess Al? 2.13
How many grams of O2 are needed to react with 100 grams of Al? 88.9
How many grams of Al2O3 can be produced from 75 grams of Al and excess O2? 142
2. For the reaction:
2 B4H10 + 11 O2  4 B2O3 + 10 H2O
a.
b.
c.
d.
How many moles of O2 are needed to react with 3.7 moles of B4H10? 20.35
How many moles of B2O3 can be made from 0.075 moles of O2 and excess B4H10? 0.0273
How many grams of B4H10 are needed to react with 0.25 grams of O2? 0.0756
How many grams of water will be made from 0.43 grams of B4H10 and excess O2? 0.727
3. For the reaction:
Na2SO4 (s) + 4 C (s)  Na2S (s) + 4 CO (g)
a. If you mix 15 g of Na2SO4 with 7.5 g of C, how many grams of Na2S can be made? 8.25 g
b. If you mix 10 g of Na2SO4 with 14.2 g of C, how many grams of CO can be made? 5.50 g
4. For the reaction:
2 Al + 3 Cl2  2 AlCl3
a. If you mix 2.7 g Al and 4.05 g Cl2, which reactant is limiting? Cl2
b. What mass of AlCl3 can be made? 5.07 g
c. What mass of the excess reactant remains? 1.674 g of Al leftover
5. For the reaction:
CaO + 2 NH4Cl  2 NH3 + H2O + CaCl2
a. If 112 g of CaO and 224 g of NH4Cl are mixed, what is the theoretical yield of NH3? 67.9
g
b. If only 16.3 g of NH3 is actually made, what is the % yield? 24.0%
6. For the reaction:
CuSO4 + 4 NH3  Cu(NH3)4SO4
a. If you react 25 mL of 1.25 NH3 with 7.5 grams of CuSO4, what is the theoretical yield of
the product in grams? 1.78 g
b. If you only get 0.63 grams of the product, what is the % yield? 35.4%
7. For the reaction:
2 NaN3  2 Na + 3 N2
What mass of NaN3 is required to give 75 L of N2 at a pressure of 1.3 atm and 25 oC? 172.9 g
8. For the reaction:
2 C8H18 + 25 O2  16 CO2 + 18 H2O
a. If 0.095 g of C8H18 burns in O2, what will be the pressure of water vapor in a 4.75 L flask
at 30 oC? 0.0393 atm
b. If the O2 gas needed for complete combustion was in the 4.75 L flask at 22 oC, what would
its pressure have been? 0.0531 atm
9. For the reaction:
4 KO2 + 2 CO2  2 K2CO3 + 3 O2
What mass of KO2 is required to react with 8.90 L of CO2 at 30 oC and 780 mm Hg? 52.2 g
10. For the reaction:
Na2CO3 + 2 HCl  2 NaCl + CO2 + H2O
If 38.55 mL of HCl is needed to titrate 2.150 g of Na2CO3, what is the M of HCl? 1.05 M
11. For the reaction:
H2A + 2 NaOH  Na2A + 2 H2O
If 36.04 mL of 0.509 M NaOH is needed to reach the endpoint of 0.954 g of H2A, what is the
molar mass of H2A? 104 g/mole
Chemical Analysis
1. A mixture of CuSO4 and CuSO4*5H2O has a mass of 1.245 g. After heating to drive off the
water, the mass is only 0.832 g. What was the mass % of CuSO4 and CuSO4*5H2O in the
original mixture?
1.245 g – 0.832 g = 0.413 g H2O lost
0.413 g H2O (1 mole/18 g) = 0.02294 moles H2O
Since CuSO4*5 H2O has 5 waters in it = 0.004589 moles CuSO4*5H2O
0.004589 moles CuSO4*5H2O (249.6 g/mole) = 1.1454 g CuSO4*5 H2O
1.145 g CuSO4*5 H2O/1.245 g total *100% = 92% CuSO4*5 H2O and 8% CuSO4
2. For the reaction:
2 NaHCO3  Na2CO3 + CO2 + H2O
Heating a 1.7184 g sample of impure NaHCO3 gives 0.196 g of CO2. What was the mass
percent of NaHCO3 in the original sample?
0.196 g CO2 *(1 mole/44 g) = 0.00445 moles CO2
0.00445 moles CO2 *(2 moles NaHCO3/1 mole CO2) = 0.00891 moles NaHCO3
0.00891 moles NaHCO3 *(84 g/mole) = 0.748 g NaHCO3
0.748 g NaHCO3/1.7184 g total *100% = 43.55%
3. For the reaction:
BeC2O4 *3 H2O (s)  BeC2O4 (s) + 3 H2O (g)
a. If 3.21 g of BeC2O4*3H2O is heated, calculate:
i) the mass of BeC2O4 (s) formed
3.21 g BeC2O4*3 H2O *(1 mole/151 g) = 0.02126 moles BeC2O4*3H2O
0.02126 moles BeC2O4*3H2O = 0.02126 moles BeC2O4
0.02126 moles BeC2O4*(97 g/mole) = 2.06 g BeC2O4
ii) the volume of H2O (g) released at 220 oC and 735 mm Hg
0.02126 moles BeC2O4*3H2O *(3 moles H2O/1 mole BeC2O4*3H2O) = 0.06378 moles
V = nRT/P = (0.06378 moles)(0.0821)(493 K)/(0.967 atm) = 2.67 L
b. A 0.345 g of BeC2O4 which contained an inert impurity was dissolved in enough water to
give 100 mL of solution. A 20 mL portion of this was titrated with 0.0150 M KMnO4 and
required 17.80 mL to reach the endpoint. The equation is as follows:
16 H+ + 2 MnO4-1 + 5 C2O4-2  2 Mn+2 + 10 CO2 + 8 H2O
i) Identify the reducing agent
C2O4-2
ii) Calculate the number of moles of each of the following at the endpoint of the titration:
 MnO4-1 (0.0178 L)(0.015 M) = 2.67x10-4 moles MnO4-1
 C2O4-2 (2.67x10-4 moles MnO4-1)(5 moles C2O4-2/2 moles MnO4-1) = 6.675x10-4 moles
iii) Calculate the total moles of C2O4-2 that were in the 100 mL sample
6.675x10-4 moles in the 20 mL portion means 6.675x10-4 moles *5 = 0.00334 moles total in 100 mL
iv) Calculate the mass % of BeC2O4 in the impure 0.345 g sample
0.00334 moles BeC2O4 *(97 g/mole) = 0.324 g
0.324 g/0.345 g total*100% = 93.9%
4. For the following reaction:
Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g)
A 16.2 L sample of CO (g) at 1.5 atm and 200 oC is combined with 15.39 g of Fe2O3
a. How many moles of CO are present?
N = PV/RT = (1.5 atm)(16.2 L)/(0.0821)(473 K) = 0.6258 moles CO
b. What is the limiting reagent? Prove it with calculations.
0.0964 moles Fe2O3 & 0.6258 moles CO
0.0964 moles Fe2O3 *(3 moles CO/1mole Fe2O3) = 0.289 moles CO needed
Fe2O3 is the limiting reagent
c. How many grams of Fe are made?
0.0964 moles Fe2O3 *(2 moles Fe/1mole Fe2O3) = 0.1928 moles Fe
0.1928 moles Fe *(55.85 g/mole) = 10.77 g Fe
d. How many L of CO2 are made?
0.0964 moles Fe2O3 *(3 moles CO2/1mole Fe2O3) = 0.2892 moles CO2
V = nRT/P = (0.2892 moles)(0.0821)(473 K)/(1.5 atm) = 7.49 L
5. For the reaction:
3 Ba(NO3)2 (aq) + 2 H3PO4 (aq)  Ba3(PO4)2 (s) + 6 HNO3 (aq)
If 200 mL of 3 M Ba(NO3)2 are mixed with 300 mL of 1 M H3PO4, calculate the following:
a) The mass of Ba3(PO4)2 (s) formed
3 M Ba(NO3)2 (200/500) = 1.2 M Ba(NO3)2 *(0.5 L total) = 0.60 moles
1 M H3PO4 (300/500) = 0.6 M H3PO4 *(0.5 L total) = 0.30 moles
3 Ba(NO3)2 (aq) + 2 H3PO4 (aq)  Ba3(PO4)2 (s) + 6 HNO3 (aq)
Start 0.60 moles
0.30 moles
Change 0.45 moles
0.30 moles
0.15 moles gained
End 0.15 moles
0 moles left
Gained 0.15 moles Ba3(PO4)2 *(602.2 g/mole) = 90.3 g
b) The final concentration of the NO3-1 ion in solution
3 M Ba(NO3)2 means 6 M NO3-1 but is diluted by a factor of (200/500) means 2.4 M
6. For the reaction:
MnO4-1 + 5 Fe+2 + 8 H+  Mn+2 + 5 Fe+3 + 4 H2O
A 0.598 g sample of an iron-containing compound was analyzed. It required 22.25 mL of
0.0123 M KMnO4 to reach the equivalence point. What is the mass % of iron in the
sample?
(0.02225 L MnO4-1)(0.0123 M) = 2.74x10-4 moles MnO4-1
2.74x10-4 moles MnO4-1 *(5 moles Fe/1 mole MnO4-1) = 0.001368 moles Fe
0.001368 moles Fe *(55.85 g Fe/1mole Fe) = 0.0764 g Fe
0.0764 g Fe/0.598 g total *100% = 12/8%