Stoichiometry
2002
Question 11 (a)
(i) In what type of household product would you expect to find sodium hypochlorite?
(4)
(ii) A solution of sodium hypochlorite, NaOCl, is labelled as having a concentration of 5% (w/v).
Express the concentration of the sodium hypochlorite solution in grams per litre.
(6)
100 cm3 of this 5% (w/v) solution were reacted with excess chloride ion and acid according to the equation.
OCl- + Cl- + 2H+ → Cl2 + H2O
(iii) How many molecules of chlorine gas were liberated?
(9)
(iv) What volume would this quantity of chlorine gas occupy at s.t.p.?
(6)
2002
(a)
Question 11.
(i)
(ii)
(iii)
(iv)
2004
bleach / mould (mildew) remover /steriliser / disinfectant
50
5 % (w/v) = 5g per 100 cm3 (3) x 10 = 50 g l–1 (3)
4 x 1022
23 = 4 x 1022 (3)
2 (3) x 6 x 10
*Using formula mass of OCl¯ (–3) Rounding off to 0.06 mol (–1)
1.5 l / 1500 – 1503 cm3
0.067 x 22.4 = 1.5 (6)
Rounding off to 0.06 not penalised a second time. Note: accept answers given to one significant figure
(4)
(6)
(9)
(6)
Question 10 (a)
Hydrochloric acid is severely corrosive to skin and eyes and toxic by inhalation or ingestion. It should be
handled carefully and stored safely. The entire contents of a bottle containing 2.5 litres of concentrated
hydrochloric acid were accidentally spilled in a laboratory. The spilled acid was neutralised by adding solid
powdered sodium carbonate. The neutralisation reaction is described by the following equation.
Na2CO3 + 2HCl = 2NaCl + H2O + CO2
The spilled acid was a 36% (w/v) solution of hydrogen chloride in water.
(i) Calculate the number of moles of hydrochloric acid spilled.
(10)
(ii) What was the minimum mass of anhydrous sodium carbonate required to completely neutralise
all of the spilled hydrochloric acid?
(9)
(iii) What volume of carbon dioxide in litres, at room temperature and pressure, was produced in this
neutralisation reaction?
(6)
2004
(a)
Question 10
(i)
(ii)
(iii)
2004
MOLES: Allow 24.60 to 25.00 mol
36% (w/v) = 36 g in 100 cm3 = 360g l-1
2.5 l contains 2.5 x 360 = 900g
Moles = mass/ Mr HCl = 900 / 36.5 = 24.66 (4 + 2 x 3)
Allow 1303.80 to 1325.00 g
24.60 to 25.00 mol HCl ≡ 12.30 to 12.50 mol Na2CO3 (3)
12.30 to 12.50 x 106* = 1303.80 to 1325.00 (6)
[*106 essential unless calculation shown e.g. = 2 x 23 + 12 + 3 x 16 = 107 (slip)]
Allow 295.20 to 300.00
24.60 to 25.00 mol HCl  12.30 to 12.50 mol CO2 (3)
12.30 to 12.50 x 24 = 295.20 to 300.00 (3)
[Give (0) if 22.4 used]
(10)
(9)
(6)
Question 10 (c)
State Avogadro’s law.
(i) What is an ideal gas?
(ii) State one reason why ammonia gas deviates from ideal gas behaviour.
(iii) A small quantity of the volatile organic solvent propanone (C3H6O) evaporates at room temperature
and pressure. Use the equation of state for an ideal gas to calculate the volume, in litres, of propanone
vapour formed when 0.29 g of liquid propanone evaporates taking room temperature as 20 ºC and room
pressure as 101 kPa.
(5)
(5)
(3)
(12)
2004
Question 10.
(c)
(i)
(ii)
(iii)
2005
STATE: equal volumes of gases contain equal numbers of molecules (particles) (3)
under same conditions of temperature and pressure (2)
[Allow ‘1 mole of a gas occupies 22.4 l at s.t.p. for 3 marks only’]
perfectly obeys the gas laws (Boyle’s law, the kinetic theory, PV = nRT) (3)
under all conditions of temperature and pressure (2)
polar / intermolecular forces (or named intermolecular force) / volume of molecules not
negligible) / collisions not perfectly elastic
ANY ONE:
0.12 litres
PV = nRT => V = nRT
T = 20 + 273 = 293 K
P
P = 101 x 1000 = 1.01 x 105 Pa
n = 0.29 = 0.005 mol (3)
58
V = 0.005 x 8.3 x 293 (3)
= 0.00012 m3
(3)
5
1.01 x 10
0.00012 x 1000 = 0.12 litres (3)
(5)
(5)
(3)
(12)
Question 10(a)
An indigestion tablet contains a mass of 0.30 g of magnesium hydroxide [Mg(OH)2] as its only basic
ingredient. The balanced chemical equation for the reaction between magnesium hydroxide and hydrochloric
acid (HCl(aq)), the acid produced in the stomach, is as follows:
Mg(OH)2 + 2HCl = MgCl2 + 2H2O
(i) Calculate the volume of 1.0 M HCl neutralised by two of these indigestion tablets.
Give your answer correct to the nearest cm3.
(8)
(ii) What mass of salt is formed in this neutralisation?
(5)
(iii) How many magnesium ions are present in this amount of the salt?
(6)
(iv) Another indigestion remedy consists of a suspension of magnesium hydroxide [Mg(OH)2] in water
and is marked 6% (w/v). What volume of this second indigestion remedy would have the same neutralising
effect on stomach acid as two of the indigestion tablets mentioned earlier?
(6)
2005
(a)
Question 10
(i)
(ii)
(iii)
(iv)
2005
CALC: 20 / 21 cm3 [(–1) for answer not given to nearest cm3 or given in litres]
2 x 0.3 = 0.6 g  58*  0.01 / 0.0103 mol (3)
0.01 / 0.0103 mol Mg(OH)2 ≡ 0.02 / 0.021 mol HCl (3)
volume of 1.0M HCl = 20 / 21 cm3 (2)
CALC: 0.95 – 0.99 g [correct to second decimal place]
0.01 / 0.0103 mol Mg(OH)2 ≡ 0.01 / 0.0103 mol MgCl2 (3)
0.01 / 0.0103 mol MgCl2 x 95*  0.95 – 0.98 g (2)
[* addition must be shown for error to be treated as slip]
CALC: 6 x 1021 – 6.21 x 1021
0.01 / 0.0103 mol MgCl2 ≡ 0.01 / 0.0103 mol Mg2+ (3)
0.01 / 0.0103 x 6 x 1023 = 6 x 1021 – 6.21 x 1021 (3)
HOW: 10 cm3
6 % (w/v) = 6 g in 100 cm3 (3) => 0.6 g in 10 cm3 (3)
(8)
(5)
(6)
(6)
Question 11 (b)
(i) Define a mole of a substance.
(7)
(ii) State Avogadro’s law.
(6)
(iii) A foil balloon has a capacity of 10 litres. How many atoms of helium occupy this balloon when it is
filled with a 10% (v/v) mixture of helium in air at room temperature and pressure?
(12)
2005
(b)
Question 11
(i)
(ii)
(iii)
DEFINE: has same number of particles* as 12 g of carbon-12 OR contains the Avogadro
number (L, 6 x 1023) of particles* OR relative molecular mass in grams (g)
[Accept “atoms”, “ions” or “molecules” in place of “particles”]
STATE: equal volumes of gases contain equal numbers of molecules (moles) (3)
under same conditions of temperature and pressure (3) (Do not accept "at s.t.p.")
[Allow 3 marks for “one mole of a gas at s.t.p. occupies 22.4 litres”]
HOW MANY: 2.4 x 1022 – 2.5 x 1022 atoms
10 % (v/v) = 10 cm3 per 100 cm3 / 10 litres per 100 litres (3)
=> volume of helium = 1 litre (3)
= 0.0416٠mol (3) x 6 x 1023 = 2.4 x 1022 – 2.5 x 1022 atoms (3)
(7)
(6)
(12)
2006
Question 11 (a)
(i) What is an ideal gas?
(4)
(ii) Give one reason why a real gas like carbon dioxide deviates from ideal behaviour.
(3)
(iii) Assuming ideal behaviour, how many moles of carbon dioxide are present in 720 cm3 of the
gas at 10 ºC and a pressure of 1 × 105 Pa? Give your answer correct to one significant figure.
(9)
(iv) How many molecules of carbon dioxide are present in this quantity of carbon dioxide?
(3)
(v) The reaction between carbon dioxide and limewater is represented by the following balanced equation.
Ca(OH)2 + CO2  CaCO3 + H2O
What mass of calcium hydroxide is required to react completely with the quantity of carbon dioxide gas
given in (iii) above?
(6)
2006
(a)
Question 11
(i)
(ii)
(iii)
(iv)
(v)
2007
WHAT: perfectly obeys the gas laws (Boyle’s law, kinetic theory, PV = nRT) under
all conditions of temperature and pressure
GIVE: intermolecular forces (attractions between molecules, named correct inter-molecular
force) / molecules have volume (molecules take up space, volume of molecules not
negligible) / collisions not perfectly elastic
ANY ONE:
MOLES: 0.03 mol
PV = nRT
1 x 105 x 720 x 10-6 = n x 8.3 x 283 (2 x 3)
n = 0.03 (3) [Not given correct to one significant figure (–1)]
MOLECULES: 1.8 x 1022
0.03 x 6 x 1023 = 1.8 x 1022 (3)
MASS: 2.22 g
0.03 mol CO2 ≡ 0.03 mol Ca(OH)2 (3)** 0.03 x 74* = 2.22 (3) [* Addition must be shown for error to
be treated as a slip.] ** Can be given for 1 : 1 ratio or for 0.03 mol Ca(OH)2
(4)
(3)
(9)
(3)
(6)
Question 10 (b)
(i) State Avogadro’s law.
(7)
(ii) Carbon dioxide is stored under pressure in liquid form in a fire extinguisher. Two kilograms of
carbon dioxide are released into the air as a gas on the discharge of the fire extinguisher.
What volume does this gas occupy at a pressure of 1.01 × 105 Pa and a temperature of 290 K? (9)
What mass of helium gas would occupy the same volume at the same temperature and pressure? (6)
(iii) Give one reason why carbon dioxide is more easily liquefied than helium.
(3)
2007
(b)
Question 10
(i)
(ii)
(iii)
2008
STATE: equal volumes of gases contain equal numbers of molecules ( moles) (4)
under same conditions* of temperature and pressure (3) [* Do not accept “under all conditions”.]
[Do not accept “at s.t.p.”] [Allow (3) for “the molar volume at s.t.p. is 22.4 litres.]
WHAT 1.069 – 1.10 m3 [Accept 1.1 but not greater]
2000 ÷ 44* = 45.4 / 45.5 mol (3) [*addition must be shown for error to be treated as slip.]
V = nRT = 45.4/45.5 x 8.3 x 290 [or other correct form] (3)
P
1.01 x 105
= 1.069 – 1.10 (3)
WHAT: 0.182 kg / 182 g [or answers rounding off to these figures]
45.4 / 45.5 x 4 (3) = 182 g / 0.182 kg (3)
GIVE: stronger intermolecular (Van der Waals’, dipole-dipole) forces (attractions) / higher
mass / bigger molecules / polarity of C to O bond / has more electrons
[To allow
opposite points Helium must be mentioned.]
Question 11 (b)
From July 2008 changes will apply to the way in which taxes are levied on new cars bought
in Ireland. Vehicles that, in controlled tests, have higher levels of carbon dioxide emission
per kilometre travelled will be subject to higher levels of taxation. The measures are designed
to encourage the purchase of cars that are more fuel-efficient and have lower CO2 emissions.
(7)
(9)
(6)
(3)
The manufacturer’s specification for a certain diesel-engined car is 143 g CO2 / km (i.e. the car produces 143 g
of CO2 for every kilometre travelled).
The car is used for morning and afternoon school runs totalling 8 km per day.
Use the manufacturer’s CO2 emission figure to calculate the amount of CO2 produced each day during the school
runs in terms of
(i) the mass of CO2,
(ii) the number of moles of CO2,
(iii) the volume of CO2 at room temperature and pressure.
(18)
If a large SUV (sports utility vehicle) with a CO2 emission rating of 264 g CO2 / km were used instead of the car
mentioned above, how many more litres of CO2 would be released into the atmosphere per day during the school
runs?
(7)
2008
Question 11 (b)
(b) (i) MASS: 1144 g
144 X 8
(6)
(ii) MOLES: 26 mol 1144 divided 44 (3) = 26 (3) To be accepted as a slip, some work must be shown in these
calculations
(6)
.
(iii) VOLUME: 624 l 26 X 24 (3) = 624 (3)
SUV: 528 l
264 x 8 ÷ 44 x 24 = 1152 (4)
1152 – 624 = 528 (3)
264 – 143 = 121 (3)
121 x 8 ÷ 44 x 24 = 528 (4)
(6)
(7)
2112 – 1144 = 968 (3)
968 ÷ 44 x 24 = 528 (4)
48 – 26 = 22 (3)
22 x 24 = 528 (4)
[Note: subtraction step (3); other step(s) (4)]
[In part (iii), using 22.4 for 24 loses the 3 (4) marks for that step but the candidate is penalised once only. The
same applies to the use of PV = nRT except in cases where the correct answer is obtained.]
2009
Question 10 (a)
State Avogadro’s law.
Give two assumptions of the kinetic theory of gases.
Give two reasons why real gases deviate from ideal gas behaviour.
How many moles of gas are present in a sample containing 1.8x1024 atoms of chlorine at s.t.p.?
2009
(a)
2011
(7)
(6)
(6)
(6)
Question 10
STATE: equal volumes of gases contain equal numbers of molecules (moles) (4)
under same conditions* / at same temperature and pressure (3)
[Allow (3) for ‘the molar volume at s.t.p. is 22.4 litres’.]
[*Do not accept ‘under all conditions’ or ‘at s.t.p.’ or ‘at constant temp & pressure’]
GIVE: gases made up of particles in rapid, random, straight-line motion //
volume of particles negligible // no forces of attraction or repulsion between particles // collisions
between molecules perfectly elastic (involve no energy loss) // average kinetic energy of
molecules proportional to Kelvin temperature
ANY TWO: (2 x 3)
DEVIATE: have intermolecular forces or named correct intermolecular force) // molecules
volume not negligible) // collisions not perfectly elastic
ANY TWO: (2 x 3)
HOW: 1.5 mol (6)
1.8 x 1024 ÷ 2 = 9 x 1023 molecules (3) ÷ 6 x 1023 = 1.5 moles (3)
(7)
(6)
(6)
(6)