KEY WS Gases Free Response Practice KEY
[KEY]WS Gases Free Response Practice[KEY]
Question 1 (1972)
A 5.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0 molar HCl solution.
(a)A 249 milliliter sample of dry CO2 gas, measured at 22C and 740 torr, is obtained from the reaction. How many moles of CO2 gas was produced by the reaction?
P V = n R T
(740)(0.249 L) = n (62.36)(273 + 22)
n = 0.0100 mol CO2
(b)Assuming potassium carbonate was the only substance in the mixture that reacted with the HCl solution, then the reaction below shows the production of CO2 gas from the mixture.
K2CO3(s) + 2 HCl(aq) 2 KCl(aq) + H2O(l) + CO2(g)
How many grams of K2CO3 were reacted to produce the moles of CO2(g) determined in part (a)?
0.0100 mol CO2 x 1 mol K2CO3 x 138.21 g K2CO3 = 1.38 g K2CO3
1 mol CO2 1 mol K2CO3
(c)What is the percentage of potassium carbonate in the original 5.00 gram mixture?
1.38 g K2CO3 x 100% = 27.6% K2CO3
5.00 g mix
Question 2 (1973)
A 6.19 gram sample of PCl5 is placed in an evacuated 2.00 liter flask and is completely vaporized at 252C.
(a)Calculate the pressure in mmHg in the flask if no chemical reaction were to occur.
6.19 g PCl5 x 1 mol PCl5 = 0.0297 mol PCl5
208.22 g PCl5
P V = n R T
P (2.00 L) = (0.0297 mol) (62.36) (273 + 252)
P = 486 mmHg
Question 3 (1993)
Observations about real gases can be explained at the molecular level according to the kinetic molecular theory of gases and ideas about intermolecular forces. Explain how each of the following observations can be interpreted according to these concepts, including how the observation supports the correctness of these theories. (you may use drawings or diagrams to support your explanations)
(a)When a gas-filled balloon is cooled, it shrinks in volume; this occurs no matter what gas is originally placed in the balloon.
Reducing the temperature of a gas reduces the average kinetic energy of the gas molecules. This would reduce the number (or frequency) of collisions of gas molecules with the surface of the balloon; Fewer collisions with the inside surface of the balloon would cause the internal pressure to be less than the external pressure so the balloon would shrink in volume until the internal and external pressure were equal again
(b)When the balloon described in part (a) is cooled further, the volume does not become zero; rather, the gas becomes a liquid or solid.
The molecules of the gas do have volume, but the volume of each particle is typically so small that it’s insignificant (or 0) when compared to the average distance between gas molecules at normal (not low) temperatures. When the molecules are cooled sufficiently, the intermolecular forces of attraction between them cause them to form liquids or solids of a certain non-zero volume.
(c)A flag waves in the wind.
The wind is moving molecules of air that are going mostly in one direction. Upon colliding with a solid matter surface (flag), they transfer some of their energy to it and cause it to move (flap).
Question 4 (1995)
Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking.
(a)Write a balanced equation for the complete combustion of propane gas yielding CO2(g) and H2O(l).
C3H8 + 5 O2 3 CO2 + 4 H2O
(b)Calculate the volume of O2 at 30C and 1.00 atm that is needed to burn completely 10.0 grams of propane.
10.0 g C3H8 x 1 mol C3H8 x 5 mol O2 = 1.13 mol O2
44.09 g C3H8 1 mol C3H8
P V = n R T
(1.00 atm) V = (1.13 mol) (0.08206) (273 + 30)
V = 28.1 L O2
(c)Assuming that air is 21.0 percent O2 by volume, calculate the volume of air needed for this reaction.
21.0% of air is O2
21.0% x Vair = VO2
0.210 x Vair = 28.1 L
Vair = 28.1 = 134 L air
0.210
Question 5 (1996)
Represented above are five identical balloons, each filled to the same volume at 25C and 1.0 atm of pressure with the pure gases indicated.
(a)Which balloon contains the greatest mass of gas? Explain.
CO2 because they all contain the same number of particles (same n if same T, P, and V), therefore, the most massive molecule, CO2 (molar mass = 44.01 g/mol), will have the greatest total mass.
(b)Compare the average kinetic energies (KEavg) of the gas molecules in the balloons. Justify your response by referencing principles from the kinetic-molecular theory of gases.
All have the same KEavg because temperature is directly proportional to KEavg , and all gases are at the same temperature.
Question 6 (2003)
A 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).
(c)Write the balanced equation for the reaction.
2 NO + O2 2 NO2
(d)Determine the limiting reactant. Justify your answer with a calculation.
0.176 mol NO x 2 mol NO2 = 0.176 mol NO2producedNO is limiting
2 mol NO
0.176 mol O2 x 2 mol NO2 = 0.352 mol NO2produced
1 mol O2
(e)Determine the number of moles of each gas that remain in the cylinder after the reaction is complete.
0.176 mol NO x 2 mol NO2 = 0.176 mol NO2produced (as product)
2 mol NO
0.176 mol NO x 1 mol O2 = 0.0880 mol O2consumed (as reactant)
2 mol NO
(0.176 mol O2 initial) – (0.0880 mol O2 consumed) = 0.0880 mol O2remain (as excess reactant)
(0.176 mol NO initial) – (0.176 mol NO consumed) = 0mol NOremain (as limiting reactant)
[OR]
2 NO + O2 2 NO2
Initial 0.176 mol 0.176 mol 0 mol
Change –0.176 –0.0880 +0.176 mol
End 0 mol 0.0880 mol 0.176 mol
limiting excess
(f)Calculate the total pressure, in atm, in the cylinder at 298 K after the reaction is complete.
0.176 mol NO2 + 0.0880 mol O2 = 0.264 total moles of gas remainin the cylinder
Pt V = n R T
Pt (5.00 L) = (0.264 mol) (0.08206) (298 K)
Pt = 1.29 atm
[OR]
use PV=nRT to calculate P of NO2 and O2 from moles of NO2 (nNO2) and O2 (nO2), then add the two pressures for the total pressure
Question 7 (2003)
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.
24.5 g N2 x 1 mol N2 = 0.874 mol N2
28.02 g N2(0.874 mol N2) + (0.875 mol O2)
= 1.749 total moles of gas in the cylinder
28.0 g O2 x 1 mol O2 = 0.875 mol O2
32.00 g O2
Pt V = n R T
Pt (5.00 L) = (1.749 mol) (0.08206) (298 K)
Pt = 8.55 atm
[OR]
use PV=nRT to calculate PN2 and PO2 from moles of N2 (nN2) and O2 (nO2), then add PN2 + PO2
Question 8 (2004)
Answer the following questions about carbon monoxide, CO(g), and carbon dioxide, CO2(g). Assume that both gases exhibit ideal behavior.
(a)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.
(b)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.
Both have the same KEavg because temperature is directly proportional to KEavg , and both gases are at the same temperature.
(ii)Indicate whether the number of CO2(g) molecules is greater than, equal, or less than the number of CO(g) molecules. Justify your answer.
Less because volume (V) and number of molecules (n) of gas are directly proportional (Avogadro’s law). The pressure of CO2 is half the pressure of the CO at the same temperature (T) and pressure (P) , so the flask of CO2 must contain half as many molecules.
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