CHEM1001
2014-J-4
June 2014
• Hydrazine (N2H4) reacts with dinitrogen tetraoxide (N2O4) to produce nitrogen and
water, all in the gas phase, according to the following unbalanced equation.
N2H4(g) + N2O4(g) → N2(g) + H2O(g)
Balance the above equation.
2N2H4(g) + N2O4(g) → 3N2(g) + 4H2O(g)
Describe the physical characteristics of a gas and sketch how the atoms of gaseous
nitrogen might be represented in a container.
Diatomic molecules of N2 fill the container, but
the individual molecules are far apart. The
molecules are travelling in different directions
and at different speeds colliding with each other
and the walls of the container.
1.00 L of hydrazine was mixed with 1.00 L of dinitrogen tetraoxide at 25 °C and
1.00 atm pressure. Briefly explain Avogadro's Law and determine the mole ratio of
hydrazine to dinitrogen tetraoxide present at room temperature.
Avogadro’s law states that equal volumes of gases at the same temperature and
pressure contain equal numbers of molecules.
From this, 1.00 L of hydrazine and 1.00 L of dintrogen tetraoxide contain the
same number of molecules. Hence the mole ratio of N2H4 to N2O4 is 1:1.
Using the ideal gas equation, calculate the number of moles of hydrazine gas under
these conditions.
From the ideal gas law, PV = nRT. Hence the number of moles is given by:
n = PV / RT
With P = 1.00 atm, V = 1.00 L, R = 0.08206 L atm K-1 mol-1 and T = 25 oC = 298
K:
n = (1.00 atm × 1.00 L) / (0.08206 L atm K-1 mol-1 × 298 K) = 0.0409 mol
Answer: 0.0409 mol
THIS QUESTION IS CONTINUED ON THE NEXT PAGE.
22/01(a)
Marks
7
CHEM1001
2013-J-4
June 2013
• In an experiment, 5.0 g of magnesium was dissolved in excess hydrochloric acid to
give magnesium ions and hydrogen gas. Write a balanced equation for the reaction
that occurred.
Mg(s) + 2H+(aq) à Mg2+(aq) + H2(g)
What amount of hydrogen gas (in mol) is produced in the reaction?
The molar mass of Mg is 24.31 g mol-1. 5.0 g therefore corresponds to:
number of moles = mass / molar mass = 5.0 g / 24.31 g mol-1 = 0.21 mol
From the chemical equation, each mol of Mg that reacts will give one mol of H2.
Hence,
number of moles of H2 = 0.21 mol.
Answer: 0.21 mol
What volume would the hydrogen occupy at 25 °C and 100.0 kPa pressure?
Using the ideal gas law, PV = nRT. Hence,
V = nRT / P
= (0.21 mol) × (8.314 J K-1 mol-1) × ((25 + 273) K) / (100.0 × 103 Pa)
= 0.0051 m3
As 1 m3 = 1000 L, this corresponds to:
V = 1000 × 0.0051 L = 5.1 L
Answer: 5.1 L
22/01(a)
Marks
4
CHEM1001
2013-J-9
June 2013
• A certain mixture of gases containing 0.24 mol of He, 0.53 mol of N2 and 0.05 mol of
Ne is placed in a container with a piston that maintains it at a total pressure of 1.0 atm.
This gas mixture is now heated from its initial temperature of 290 K to 370 K by
passing 2.08 kJ of energy into it.
Calculate the volume occupied by the gas at 370 K.
The total number of moles of gas = (0.24 + 0.53 + 0.05) mol = 0.82 mol.
Using the ideal gas law, PV = nRT, the volume occupied is:
V = nRT / P
= (0.82 mol) × (0.08206 L atm K–1 mol–1) × (370 K) / (1.0 atm)
= 25 L
Answer: 25 L
22/01(a)
Marks
2
CHEM1001
2012-J-7
June 2012
• The equation for the detonation of nitroglycerine, C3H5N3O9(l), is given below.
4C3H5N3O9(l) → 6N2(g) + 12CO2(g) + 10H2O(g) + O2(g)
What mass of nitroglycerine is required to produce 720 L of product gases at
1800 °C and 1.00 atm? Assume all gases behave as ideal gases. Show all working.
Using the ideal gas equation, PV = nRT, the total number of moles of gas
produced is:
n = PV / RT = (1.00 atm × 720 L) / (0.08206 atm L K-1 mol-1 × (1800 + 273) K)
= 4.23 mol
From the chemical equation, detonation of 4 mol of nitroglycerine gives (6 + 12 +
10 + 1) mol = 29 mol of gases. Therefore:
number of moles of nitroglycerine required = (4/29) × 4.23 mol = 0.584 mol
The molar mass of nitroglycerine is:
molar mass = [3 × 12.01 + 5 × 1.008 (H) + 3 × 14.01 (N) + 9 × 16.00 (O)] g mol-1
= 227.1 g mol-1
The mass of nitroglycerine is therefore:
mass = number of moles × molar mass = (0.582 mol) × (227.1 g mol-1) = 130 g
Answer: 130 g
22/01(a)
3
CHEM1001
2010-J-6
June 2010
• Respiration involves the oxidation of glucose to produce carbon dioxide, water and
energy:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l)
Balance this equation.
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
What volume of CO2(g) is produced from the oxidation of 10.0 g of glucose under
body conditions (37 °C, 1.00 × 105 Pa)?
The molar mass of glucose is:
molar mass = (6 × 12.01 + 12 × 1.008 + 6 × 16.00) g mol-1 = 180.156 g mol-1
Therefore, 10.0 g corresponds to:
number of moles of glucose = mass / molar mass
= 10.0 g / 180.156 g mol-1 = 0.0555 mol
From the balanced chemical equation, 1 mol of glucose leads to 6 mol of CO2(g)
so:
number of moles of CO2(g) generated = 6 × 0.0555 mol = 0.333 mol
From the ideal gas law, PV = nRT with R = 8.314 J K-1 mol-1 when the pressure is
used in the Pascals (the S.I. unit). The volume that this occupies is:
V = nRT / P
= (0.333 mol) × (8.314 J K-1 mol-1) × ((37 + 273) K) / (1.00 × 105 Pa)
= 8.58 × 10-3 m3 = 8.58 L
Note that as S.I. units were used, the volume is returned in m3 requiring the
conversion to litres.
Alternatively, the pressure can be converted from Pa to atm:
P = 1.00 × 105 Pa = (1.00 × 105 / 101.3 × 103) = 0.987 atm
From the ideal gas law, PV = nRT with R = 0.08206 L atm K-1 mol-1 when the
pressure is used in the atmospheres. The volume that this occupies is:
V = nRT / P
= (0.333 mol) × (0.08206 L atm K-1 mol-1) × ((37 + 273) K) / (0.987 atm)
= 8.58 L
Answer: 8.58 L
22/01(a)
Marks
4
CHEM1001
2010-J-7
June 2010
• A 2.4 g sample of zinc was dropped into 0.250 L of 5.0 M HCl in a 5.00 L container
at 25 °C with an initial pressure of 1.0 atm and then the vessel sealed. Calculate the
final pressure inside the container.
Hint: The volume occupied by the HCl is significant.
The reaction occurring is:
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
The number of moles of zinc added is:
number of moles of Zn(s) = mass / molar mass = 2.4 g / 65.39 g mol-1 = 0.037
mol
The number of moles of H+(aq) initially present is:
number of moles of H+(aq) = concentration × volume
= 5.0 mol L-1 × 0.250 L = 1.25 L
As 2 mol of H+(aq) is required for every 1 mol of Zn(s), the H+(aq) is in excess
and zinc is the limiting reagent. From the chemical equation, 1 mol of Zn(s)
produces 1 mol of H2(g) and so:
number of moles of H2(g) = 0.037 mol
This gas is contained within 4.75 L of the container as the HCl solution occupies
0.250 L. Using the ideal gas law, PV = nRT with R = 0.08206 L atm K-1 mol-1, this
amount will lead to a pressure:
P = nRT / V
= (0.037 mol) × (0.08206 L atm K-1 mol-1) × ((25 + 273) K) / (4.75 L)
= 0.19 atm
As the initial pressure was 1.0 atm, the total pressure in the container has
increased to:
total pressure = (1.0 + 0.19) atm = 1.2 atm
Answer: 1.2 atm
22/01(a)
6
CHEM1001
2009-J-7
June 2009
• Propane, C3H8, is commonly used in barbecue gas cylinders, its complete combustion
yielding water and carbon dioxide as the only products. What volume of CO2 is
produced at 0 oC and 1.0 atm from the complete combustion of 15.0 L of propane at a
pressure of 4.5 atm and a temperature of 25 oC?
Using the ideal gas law, PV = nRT, 15.0 L of propane at 25 °C ( = 298 K) and 4.5
atm corresponds to:
n=
=
. . . = 2.76 mol
The combustion reaction is:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Hence, combustion of 2.76 mol of C3H8(g) will give rise to (3 × 2.76) mol = 8.28
mol.
At 0 °C ( = 273 K) and 1.0 atm, this amount will occupy:
n=
=
. . . Answer: 190 L
= 190 L
22/01(a)
Marks
3
CHEM1001
2007-J-10
June 2007
• The Voyager I spacecraft determined that the atmospheric pressure at the surface of
Saturn’s moon, Titan, is 1.6 times that of earth and that the atmosphere contains
6.0 mol % methane, CH4. What is the partial pressure of methane on Titan in mmHg?
22/01(a)
Marks
2
If the atmospheric pressure on Titan is 1.6 times that on earth, Ptotal = 1.6 atm.
If the atmosphere contains 6.0 mol % methane, then:
PCH4 = 0.06 × 1.6 = 0.096 atm = 0.096 × 760 = 73 mmHg
Answer: 73 mmHg
• Many gases are available for use in compressed gas cylinders, in which they are
stored at high pressures. Calculate the mass of O2 that can be stored at 20. °C and
170. atm pressure in a cylinder with a volume of 60.0 L.
Using the ideal gas law, PV = nRT, the number of moles that can be stored is:
n=
PV
(170.) × (60.0)
=
= 424mol
RT (0.08206) × (20. + 273)
As the molar mass of O2 is (2 × 16.00) = 32.00, this corresponds to a mass of:
mass = number of moles × molar mass = 424 × 32.00 = 13600 g = 13.6 kg
Answer: 13,600 g or 13.6 kg
What volume would this mass of oxygen occupy at 1.00 atm pressure and 20 °C?
Using the ideal gas law, the volume of 424 mol at 1.00 atm and 20 °C is:
nRT (424) × (0.08206) × (20 + 273)
=
= 10.2 × 104 L
P
(1.00)
Equivalently, P1V1 = P2V2 gives:
V=
V2 =
P1V1 (170) × (60.0)
=
= 1.02 × 104 L
P2
(1.00)
Answer: 1.02 × 104 L
3
CHEM1001
2006-J-11
June 2006
• When “dry ice”, solid carbon dioxide, is heated to 400 K it becomes gaseous. An
88.0 g sample of solid carbon dioxide is placed into a sealed 100 L container that is
initially at a pressure of 1.00 atm and a temperature of 298 K. The container is heated
to 400 K. What will be the final pressure inside the container?
The molar mass of CO2 is (12.01 (C)) + (2 × 16.00 (O)) = 44.01. The sample of
88.0 g therefore corresponds to:
moles of CO2 =
mass
88.0
=
=2.00 mol
molar mass 44.01
The pressure due to this amount of CO2 in a container of volume 100 L at 400 K
is given by the ideal gas law, PV = nRT, as:
pCO 2 =
nRT (2.00)×(0.08206)×(400)
=
= 0.656atm
V
100
The container initially contained a gas with a pressure of 1.00 atm at 298 K. The
ideal gas equation can be applied to this gas at the two temperatures and as the
number of moles and the volume is constant, this gas will have a pressure at 400
K corresponding to:
p(T2 ) T2
400
400
=
or p(400 K) = p(298 K)×
=1.00×
=1.34atm
p(T1 ) T1
298
298
The total pressure at 400 K is therefore 0.656 + 1.34 = 2.00 atm
Answer: 2.00 atm
22/01(a)
Marks
4