Stoichiometry, Gas Laws, Lab Questions KEYS
2008 form B question #3
A 0.150 g sample of solid lead(II) nitrate is added to 125 mL of 0.100 M sodium iodide solution. Assume no change in
volume of the solution. The chemical reaction that takes place is represented by the following equation.
(a)
(b)
(c)
(d)
(e)
Pb(NO3)2(s) + 2 NaI(aq)  PbI2(s) + 2 NaNO3(aq)
List an appropriate observation that provides evidence of a chemical reaction between the two
compounds.
Calculate the number of moles of each reactant.
Identify the limiting reactant. Show calculations to support your identification.
Calculate the molar concentration of NO3–(aq) in the mixture after the reaction is complete.
Circle the diagram below that best represents the results after the mixture reacts as completely as
possible. Explain the reasoning used in making your choice.
Answer:
(a) formation of a yellow ppt
1 mol
(b) 0.150 g Pb(NO3)2 
= 4.5310-4 mol Pb(NO3)2
331.2 g
0.100 mol
0.125 mL 
= 0.0125 mol NaI
1L
(c) Pb(NO3)2; since they react in a 1:2 ratio of Pb(NO3)2:NaI, 6.2510-3 mol Pb(NO3)2 would be required to
react with all the NaI, therefore, you would run out of the lead nitrate first
4.53  10 3 mol Pb(NO 3 )2 
(d) the nitrate is a spectator ion,
(e)
0.125 L
2 mol NO-3
1 mol Pb(NO 3 )2
= 7.2510-3 M
the solution contains the sodium and nitrate spectator ions as well as some excess iodide.
2004 D
Answer the following questions about carbon monoxide, CO(g), and carbon dioxide, CO2(g). Assume that both
gases exhibit ideal behavior.
(a) Draw the complete Lewis structure (electron dot diagram) for the CO molecule and for the CO2 molecule.
(b) Identify the shape of the CO2 molecule.
(c) One of the two gases dissolves readily in water to form a solution with a pH below 7. Identify the gas and
account for this observation by writing a chemical equation.
(d) A 1.0 mol sample of CO(g) is heated at constant pressure. On the graph below, sketch the expected plot of
volume verses temperature as the gas is heated.
(e) Samples of CO(g) and CO2(g) are placed in 1 L containers at the conditions in the diagram below.
(i) Indicate whether the average kinetic energy of the CO2 is greater than, equal to, or less than the
average kinetic energy of the CO(g) molecules. Justify your answer.
(ii) Indicate whether the root-mean-square speed of the CO2(g) molecules is greater than, equal to or less
than the root-mean-square speed of the CO(g) molecules. Justify your answer.
(iii) Indicate whether the number of CO2(g) molecules is greater than, equal, or less than the number of
CO(g) molecules. Justify your answer.
Answer:
(a)
(b) linear
(c) CO2; CO2 + H2O  H+ + HCO3–
(d)
(e) (i) equal to; at the same temperature, all gas molecules have the same kinetic energy
(ii)
less, since CO2 has a molar mass of 44 and CO has a mass of 28, the lighter molecule is faster at
the same temperature
(iii) less; Avogadro’s Hypothesis, equal volumes of gas at the same temperature and pressure contain equal
number of molecules. Since the pressure of CO2 is half the pressure of the CO, it must contain half as many
molecules.
2009 part A, question #2
A student was assigned the task of determining the molar mass of an unknown gas. The student measured the
mass of a sealed 843 mL rigid flask that contained dry air. The student then flushed the flask with the unknown
gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 23˚C and 750. torr. The
data for the experiment are shown below.
Volume of sealed flask
843 mL
Mass of sealed flask and air
157.70 g
Mass of sealed flask and unknown gas
158.08 g
(a) Calculate the mass, in grams, of the dry air that was in the sealed flask. (The density of dry air is 1.18
g L-1 at 23.0˚C and 750. torr.)
(b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it).
(c) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask.
(d) Using the information above, calculate the value of the molar mass of the unknown gas.
After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide
(44.0 g mol-1).
(e) Calculate the percent error in the value of the molar mass calculated in part (d).
(f) For each of the following two possible occurrences, indicate whether it by itself could have been
responsible for the error in the student’s experimental results. You need not include any calculations
with your answer. For each of the possible occurrences, justify your answer.
Occurrence 1: The flask was incompletely flushed with CO2(g), resulting in some dry air remaining in the
flask.
Occurrence 2: The temperature of the air was 23.0˚C, but the temperature of the CO2(g) was lower than the
reported 23.0˚C.
(g) Describe the steps of a laboratory method that the student could use to verify that the volume of the
rigid flask is 843 mL at 23.0˚C. You need not include any calculations with your answer.
Answer:
1.18 g
(a) 843 mL 
= 0.995 g air
1000 mL
(b) 157.70 g flask + air
- 0.995 g air
156.71 g flask
(c)
158.08 g flask + gas
-156.71 g flask
1.37 g gas
PV

(d) n =
RT
750torr 760 (0.843 mL) 0.0343 mol
torr
atm
Lgatm
0.0821 molgK
(295.0K)
1.37 g gas / 0.0343 mol = 40.0 g mol-1
40.0  44.0   100 = -9.04% error
(e)
44.0
(f) occurrence 1: yes; dry air, with molar mass of about 28.8 g mol-1, would, if mixed with the higher molar
mass CO2 would give lower results than expected.
occurrence 2: no; if the temperature is less than expected then more of the sample would be in the flask (as
T decreases, n increases) and give too large a calculated result.
(g) measure mass of empty flask
fill with water
measure mass of flask + water
calculate mass of water, look up density of water at 23˚C
calculate volume of water
2006 D Required
5. Three pure, solid compounds labeled X, Y, and Z are placed on a lab bench with the objective of
identifying each one. It is known that the compounds (listed in random order) are KCl, Na2CO3, and
MgSO4. A student performs several tests on the compounds; the results are summarized in the table below.
Compound
pH of an Aqueous
Solution of the
Compound
Result of Adding 1.0
M NaOH to a Solution
of the Compound
Result of Adding 1.0
M HCl Dropwise to
the Solid Compound
X
Y
>7
7
No observed reaction
Evolution of a gas
No observed reaction
No observed reaction
Formation of a white
Z
7
No observed reaction
precipitate
(a) Identify each compound based on the observations recorded in the table.
Compound X__________________
Compound Y__________________
Compound Z__________________
(b) Write the chemical formula for the precipitate produced when 1.0 M NaOH is added to a solution of
compound Z.
(c) Explain why an aqueous solution of compound X has a pH value greater than 7. Write an equation as
part of your explanation.
(d) One of the testing solutions used was 1.0 M NaOH. Describe the steps for preparing 100. mL of 1.0 M
NaOH from a stock solution of 3.0 M NaOH using a 50 mL buret, a 100 mL volumetric flask, distilled
water, and a small dropper.
(e) Describe a simple laboratory test that you could use to distinguish between Na2CO3(s) and CaCO3(s). In
your description, specify how the results of the test would enable you to determine
which compound was Na2CO3(s) and which compound was CaCO3(s).
Answer:
(a) Compound X = Na2CO3
Compound Y= KCl
Compound Z = MgSO4
(b) Mg(OH)2
(c) sodium carbonate will dissolve in water to produce sodium and carbonate ions.
Na2CO3  2 Na+ + CO32–
the carbonate ions will further react with the water to produce the hydrogen
carbonate ion and a hydroxide ion, thus, an alkaline (pH>7) solution where the [OH–
] > [H+].
CO32– + H2O  HCO3– + OH–
(d) M1V1 = M2V2; (3.0 M)(V1) = (1.0 M)(100 mL); V1 = 33.3 mL
• fill the 50 mL buret with the 3.0 M NaOH solution
• dispense 33.3 mL of the solution into the 100 mL volumetric flask
• add distilled water to below line in neck of flask, mixing thoroughly
• use dropper to fill to the final mark (100.0 mL) with distilled water,
(e) place the same amount of each solid into water, Na2CO3(s) will dissolve, while CaCO3(s) will not.
2003 D Required
5. A student is instructed to determine the concentration of a solution of CoCl2 based on absorption of light
(spectrometric/colorimetric method). The student is provided with a 0.10 M solution of CoCl2 with which
to prepare standard solutions with concentrations of 0.020 M, 0.040 M, 0.060 M and 0.080 M.
(a) Describe the procedure for diluting the 0.10 M solutions to a concentration of 0.020 M using distilled water,
a 100 mL volumetric flask, and a pipet or buret. Include specific amounts where appropriate.
The student takes the 0.10 M solution and determines the percent transmittance and the absorbance at various
wavelengths. The two graphs below represent the data.
(b) Identify the optimum wavelength for the analysis.
The student measures the absorbance of the 0.020 M, 0.040 M, 0.060 M, 0.080 M and 0.10 M solutions. The
data are plotted below.
(c) The absorbance of the unknown solution is 0.275. What is the concentration of the solution?
(d) Beer’s Law is an expression that includes three factors that determine the amount of light that passes
through a solution. Identify two of these factors.
(e) The student handles the sample container (e.g., test tube or cuvette) that holds the unknown solution and
leaves fingerprints in the path of the light beam. How will this affect the calculated concentration of the
unknown? Explain your answer.
(f) Why is this method of determining the concentration of CoCl2 solution appropriate, whereas using the same
method for measuring the concentration of NaCl solution would not be appropriate?
Answer:
(a) M1V1 = M2V2; (0.10M)(V1) = (0.020M)(100. mL)
V1 = 20.0 mL
a 20.0 mL aliquot of 0.10 M solution is measured by buret or pipet. this aliquot is added to the 100-mL
volumetric flask and filled, with mixing, to the line on the neck with distilled water
(b) approx. 510 nm
(c) approx. 0.05 M
(d) extinction coefficient
path length of light
concentration of absorbing species
(e) fingerprints scatter light and the detector gets less light, the reading of absorbance is higher, indicating a
higher than expected concentration
(f) the Na+ ion does not absorb energy in the visible spectrum, whereas the Co2+ is a rose color
2003 B
A rigid 5.00 L cylinder contains 24.5 g of N2(g) and 28.0 g of O2(g)
(a) Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K.
(b) The temperature of the gas mixture in the cylinder is decreased to 280 K. Calculate each of the following.
(i) The mole fraction of N2(g) in the cylinder.
(ii) The partial pressure, in atm, of N2(g) in the cylinder.
(c) If the cylinder develops a pinhole-sized leak and some of the gaseous mixture escapes, would the
N2 (g )
ratio
in the cylinder increase, decrease, or remain the same? Justify your answer.
O2 ( g )
A different rigid 5.00 L cylinder contains 0.176 mol of NO(g) at 298 K. A 0.176 mol sample of O2(g) is added to
the cylinder, where a reaction occurs to produce NO2(g).
(d) Write the balanced equation for the reaction.
(e) Calculate the total pressure, in atm, in the cylinder at 298 K after the reaction is complete.
Answer:
1mol
(a) 24.5 g N2 
= 0.875 mol N2
28.0 g
28.0 g O2 
1mol
= 0.875 mol O2
32.0 g
L¥atm
1.75mol 0.0821 mol¥K
298K 
nRT
=
5.00L
V
= 8.56 atm
0.875 mol N 2
(b) (i)
= 0.500 mole fraction N2
1.75 mol mix
P P
P T (8.56atm)(280K)
(ii) 1  2 ; P2  1 2 
T1 T2
T1
298K
P=
= 8.05 atm  mole fraction = 8.05 atm  0.500
= 4.02 atm N2
(c) decrease; since N2 molecules are lighter than O2 they have a higher velocity and will escape more
frequently (Graham’s Law), decreasing the amount of N2 relative to O2
(d) 2 NO + O2  2 NO2
(e) all 0.176 mol of NO will react to produce 0.176 mol of NO2, only 1/2 of that amount of O2 will react, leaving
0.088 mol of O2, therefore, 0.176 + 0.088 = 0.264 mol of gas is in the container.
L¥atm
0.264 mol 0.0821 mol¥K
298 K 
nRT
P=
=
5.00L
V
= 1.29 atm
2002 D Required (repeated in thermodynamics)
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above.
The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter.
The heat released by the reaction is determined by using the equation q = mc∆T.
Assume the following.
• Both solutions are at the same temperature before they are combined.
• The densities of all the solutions are the same as that of water.
• Any heat lost to the calorimeter or to the air is negligible.
• The specific heat capacity of the combined solutions is the same as that of water.
(a) Give appropriate units for each of the terms in the equation q = mc∆T.
(b) List the measurements that must be made in order to obtain the value of q.
(c) Explain how to calculate each of the following.
(i) The number of moles of water formed during the experiment
(ii) The value of the molar enthalpy of neutralization, ∆Hneut, for the reaction between HCl(aq) and
NaOH(aq)
(d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl
and 2.0 M NaOH.
(i) Indicate whether the value of q increases, decreases, or stays the same when compared to the first
experiment. Justify your prediction.
(ii) Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut, increases, decreases, or stays
the same when compared to the first experiment. Justify your prediction.
(e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this
have on the calculated value of the molar enthalpy of neutralization, ∆Hneut? Justify your answer.
Answer:
(a) q in J, m in grams, C in J/g˚C, T in ˚C
(b) mass or volume of each solution
starting temperature of each reagent
ending temperature of mixture
(c) (i) both are 1 M acid and base and react on a 1:1 basis
1 mol HCl
1 mol H+
volume  1000 mL  1 mol HCl = mol of H+
H+ + OH–  H2O
joules released
(ii) mol H O produced
2
(d) (i) increases. Twice as much water is produced so it is twice the energy released in the same volume of
solution
twice energy
(ii) same. twice mol water = same result
(e) smaller. heat lost to the air gives a smaller amount of temperature change in the solution, which leads to a
smaller measured heat release
2000 D Required
The molar mass of an unknown solid, which is nonvolatile and a nonelectrolyte, is to be determined by the
freezing-point depression method. The pure solvent used in the experiment freezes at 10C and has a known
molal freezing-point depression constant, Kƒ. Assume that the following materials are also available.
• test tubes
• stirrer
• pipet
• stopwatch
• graph paper
• thermometer
• balance
• beaker
• ice
• hot-water bath
(a) Using the two sets of axes provided below, sketch cooling curves for (i) the pure solvent and for (ii) the
solution as each is cooled from 20C to 0.0C
Pure Solvent
Solution
(b) Information from these graphs may be used to determine the molar mass of the unknown solid.
(i) Describe the measurements that must be made to determine the molar mass of the unknown solid by
this method.
(ii) Show the setup(s) for the calculation(s) that must be performed to determine the molar mass of the
unknown solid from the experimental data.
(iii) Explain how the difference(s) between the two graphs in part (a) can be used to obtain information
needed to calculate the molar mass of the unknown solid.
(c) Suppose that during the experiment a significant but unknown amount of solvent evaporates from the test
tube. What effect would this have on (he calculated value of the molar mass of the solid (i.e., too large, too
small, or no effect)? Justify your answer.
(d) Show the setup for the calculation of the percentage error in a student’s result if the student obtains a value
of 126 g mol-1 for the molar mass of the solid when the actual value is 120. g mol-1.
Answer:
(a) (i) pure solvent
(ii) solution
(b) (i) mass of pure solvent; freezing point of pure solvent; mass of unknown solid solute added to pure
solvent; freezing point of resulting solution
(ii) determine the change in freezing point, T
mol solute
mass solute
T = Kƒ • m, where m = 1 kg of solvent and moles solute = molar mass
mass of solute•Kƒ
therefore, molar mass =
kg solvent•T
(iii) the change in temperature is the difference in the flat portions of the graph.
(c) Too small. If solvent evaporates then its mass decreases and the recorded denominator in the equation in
(b)(i) is larger than the expt. value and the resulting answer decreases.
(126g•mol-1 – 120.g•mol-1)
(d) % error =
 100%
120. g•mol-1