10.
Reacting gas volume ratios of reactants or products (Avogadro's Law,
Gay-Lussac's Law)

In the diagram above, if the
volume on the left syringe is twice that of the gas volume in the right, then
there are twice as many moles or actual molecules in the left-hand gas syringe.
-
REACTING GAS VOLUMES and MOLE
RATIO
-
Historically Gay-Lussac's Law of volumes states that 'gases combine with each other in
simple proportions by volume', but the basis of this reacting gas ratio
law, is non other than
Avogadro's Law and the 'mole concept'.
-
Avogadro's
Law states that
'equal volumes of gases at the same temperature and pressure contain the same number of molecules'
or moles of gas.
-
This means the
molecule ratio
of the equation or the relative moles of reactants and products
automatically gives us the gas volumes ratio
of reactants and products ...
-
These
calculations only apply to gaseous reactants or products AND if they are all at
the same temperature and pressure.
-
The balanced equation can
be read/interpreted in terms of either ...
-
(i) a gas volume ratio,
obviously for gaseous species only (g), AND at the same temperature and
pressure.
-
or (ii) a mole ratio,
which applies to anything in the equation, whether (g), (l) or (s).
-
Reacting gas
volume ratio calculation Example 10.1
-
Given the equation:
HCl(g)
+ NH3(g) ===> NH4Cl(s)
-
1 mole
hydrogen chloride gas combines with 1 mole of ammonia gas to give 1 mole
of ammonium chloride solid,
-
since from Avogadro's, equal
volumes of gases at the same T & P, have the same number of molecules and
equal numbers of moles have the same number of molecules, and this gives
rise to Gay-Lussac's Law of combining volumes, and we can then logically say
directly from the equation ...
-
1 volume of hydrogen chloride
will react with 1 volume of ammonia to form solid ammonium chloride
-
e.g. 25cm3 + 25cm3
==> solid product (no gas formed)
-
or 400dm3 + 400 dm3 ==>
solid product etc.
-
so, if 50 cm3 HCl
reacts, you can predict 50 cm3 of NH3 will react etc.
etc. !
-
-
-
Reacting gas
volume ratio calculation Example 10.2
-
Given the equation:
N2(g)
+ 3H2(g) ==> 2NH3(g)
-
1 mole of
nitrogen gas combines with 3 mols of hydrogen gas to form 2 mol of a
ammonia gas.
-
1 volume of nitrogen reacts with
3 volumes of hydrogen to produce 2 volumes of ammonia
-
e.g. Q: what volume
of hydrogen reacts with 50 cm3 nitrogen and what volume of
ammonia will be formed?
-
The mole ratio is 1 : 3 ==> 2,
-
so you
multiply equation ratio numbers by 50 giving ...
-
50 cm3 nitrogen
+ 150 cm3 hydrogen (3 x 50) ==>
100 cm3 of
ammonia (2 x 50)
-
-
-
Reacting gas
volume ratio calculation Example 10.3 (sometimes you need to
think 'outside the box' in some context e.g.
-
Some industrial processes produce toxic
carbon monoxide as a by-product and this gas must be dealt with.
-
One way is to burn it to release heat
energy for power generation.
-
Carbon monoxide burns in air to form
carbon dioxide according to the equation:
-
2CO(g) + O2(g)
===> 2CO2(g)
-
(a) If carbon monoxide is produced in an
industrial process at the rate of 50 dm3 per minute, what rate of
oxygen input is required to completely burn it to harmless carbon dioxide?
-
The equation reads as 2 mol CO reacts
with 1 mol O2 to form 2 mol of CO2
-
Therefore 2 volumes of CO reacts with 1
volume of O2 to form 2 volumes of CO2 by ratio.
-
Therefore volume of O2 needed
= half the volume of CO produced.
-
Therefore for every 50 dm3 of
CO produced you need 25 dm3 of oxygen to burn it.
-
Therefore you need to pump in oxygen at
the rate of 25 dm3/min
-
(b) However, it is expensive to use
oxygen, so air is employed for the combustion process. If you assume
air contains approximately 20% oxygen what rate of air needs to be pumped
into the reactor for the carbon monoxide combustion process?
-
If air is composed of 20% oxygen, then
1/5th of air is oxygen, so you need five times more air than pure oxygen.
-
Therefore rate of air needed = 5 x 25 =
125 dm3 air/min
-
-
-
Reacting
gas volume ratio calculation Example
10.4
-
Given the equation: C3H8(g) + 5O2(g)
===> 3CO2(g) + 4H2O(l)
-
Reading the balanced equation in
terms of moles (or mole ratio) ...
-
1 mole of propane gas reacts with
5 mols of oxygen gas to form 3 moles of carbon dioxide gas and 4 mols of
liquid water.
-
(a) What volume of oxygen
is required to burn 25cm3 of propane, C3H8.
-
Theoretical reactant
volume ratio is C3H8 : O2 is 1 : 5
for burning the fuel propane.
-
so actual ratio is 25
: 5x25, so 125cm3 oxygen is needed.
-
-
-
(b) What volume of carbon
dioxide is formed if 5dm3 of propane is burned?
-
Theoretical
reactant-product volume ratio is C3H8 : CO2
is 1 : 3
-
so actual ratio is 5
: 3x5, so 15 dm3 carbon dioxide is formed.
-
-
-
(c) What volume of air (1/5th
oxygen) is required to burn propane at the rate of 2dm3 per minute
in a gas fire?
-
Theoretical reactant
volume ratio is C3H8 : O2 is 1 : 5
-
so actual ratio is 2
: 5x2, so 10dm3 oxygen per minute is needed,
-
therefore, since air is
only 1/5th O2, 5 x 10 =
50 dm3
of air per minute is required.
-
-
-
Reacting
gas volume ratio calculation
Example 10.5
-
Given
the equation: 2H2(g) + O2(g) ===> 2H2O(l)
-
If 40 dm3 of
hydrogen, (at 25oC and 1 atm pressure) were burned completely
...
-
a) What volume of pure oxygen is
required for complete combustion?
-
b) What volume of air is
required if air is ~20% oxygen?
-
~20% is ~1/5,
therefore you need five times more air than pure oxygen
-
Therefore volume of air needed = 5 x
20 = 100 dm3 of air
-
-
-
c)
What
mass of water is formed?
-
The easiest way to solve this
problem is to think of the water as being formed as a gas-vapour.
-
The theoretical gas volume ratio
of reactant hydrogen to product water is 1 : 1
-
Therefore, prior to condensation
at room temperature and pressure, 40 dm3 of water vapour is
formed.
-
1 mole of gas occupies 24 dm3,
and the relative molar mass of water is 18 g/mol
-
Therefore moles of water formed
= 40/24 = 1.666 moles
-
Since moles = mass / formula
mass
-
mass = moles x formula mass
-
mass water formed = 1.666 x 18 =
30 g of H2O
-
-
-
Reacting
gas volume ratio calculation
Example 10.6
-
It was
found that exactly 10 cm3 of bromine vapour (Br2(g))
combined with exactly 30 cm3 chlorine gas (Cl2(g))
to form a bromine-chlorine compound BrClx.
-
a) From the reacting gas volume ratio,
what must be the value of x? and hence write the formula of the compound.
-
b) Write a balanced equation to
show the formation of BrClx
-
The reacting gas volume ratio is
1 : 3, therefore we can write with certainty that 1 mole (or molecule) of
bromine reacts with 3 moles (or molecules) of chlorine, and balancing the
symbol equation, results in two moles (or two molecules) of the bromine-chlorine
compound being formed in the balanced equation.
-
-
-
Reacting
gas volume ratio calculation
Example 10.7
-
Q's 10.8 and 10.9 are a couple of harder
questions, which are really two similar calculations. You would be probably
given the equations and the fraction of oxygen in air. Assume all gas volume
measurements are made at the same temperature and pressure ..
-
Reacting
gas volume ratio calculation
Example 10.8
-
What volume of oxygen/air is required
to completely burn a mixture of 10 cm3 of hydrogen and 20 cm3
of carbon monoxide?
-
(i) hydrogen: H2(g) + ½O2(g)
===> H2O(l)
-
mole ratio/gas volume ratio H2
: O2 is 1 : 0.5
-
therefore reacting volume ratio must be
the same for 10 cm3 : y cm3
-
so y = 10 x 0.5 / 1 =
5 cm3
(10/2)
-
-
-
(ii) carbon monoxide: CO(g) + ½O2(g)
===> CO2(g)
-
mole ratio/gas volume ratio CO : O2
is 1 : 0.5
-
therefore reacting volume ratio must be
the same for 20 cm3 : z cm3
-
so z = 20 x 0.5 / 1 =
10 cm3
(20/2)
-
-
-
(iii) Therefore total volume oxygen = y +
z = 5 + 10 = 15 cm3 O2
-
(iv) Assuming air is 1/5th (20%) oxygen,
the volume of air required would be 5 x 15 = 75 cm3
air
-
Reacting
gas volume ratio calculation
Example 10.9
-
(this is a definitely a bit more tricky on the equations, but check out 10.7
first)
-
What volume of oxygen/air is required
to completely burn a mixture of 20 dm3 of methane and 10 dm3
of propane?
-
(i) methane: CH4(g) + 2O2(g)
==> CO2(g) + 2H2O(l)
-
mole ratio = gas volume ratio CH4
: O2 is 1 : 2
-
therefore reacting volume ratio is 20
: y
-
so y = 20 x 2 / 1 =
40 dm3
-
(ii)
propane: C3H8(g) + 5O2(g)
==> 3CO2(g) + 4H2O(l)
-
mole ratio = gas volume ratio C3H8
: O2 is 1 : 5
-
therefore reacting volume ratio is 10
: z
-
so z = 10 x 5 / 1 =
50 dm3
-
(iii) Therefore total volume oxygen = y +
z = 40 + 50 = 90 dm3 O2
-
(iv) Assuming air is 1/5th (20%) oxygen,
the volume of air required would be 5 x 90 =
450 dm3
air
-
Reacting
gas volume ratio calculation
Example 10.10
-
This is a limiting reactant question, but
again assume all volumes are measured at the same temperature and pressure.
-
Fluorine and chlorine are both very
reactive gases, but fluorine, from the group 7 halogen reactivity trend, is
the most reactive of the two gases. Fluorine reacts with chlorine to form
chlorine(III) fluoride.
-
The balanced equation is: 3F2(g)
+ Cl2(g) ===> 2ClF3(g)
-
If 200 cm3 of fluorine gas is
reacted with 50 cm3 of chlorine gas:
-
(a) What is the theoretical gas
volume ratio for the reactants?
-
The molar ratio given by the
balanced equation above is 3 : 1,
-
so the theoretical reacting gas
volume ratio is 3 : 1
(for fluorine : chlorine)
-
-
-
(b) Which is the limiting reactant?
-
The molar ratio = the theoretical
reacting gas volume ratio of 3 : 1,
-
but the actual volume ratio of
the initial mixture is 200 : 50 or 4 : 1 (fluorine : chlorine)
-
Since 4/1 exceeds 3/1 the excess
reactant is fluorine and therefore
chlorine is the limiting
reactant.
-
-
-
(c) What volume of chlorine(III)
fluoride is formed?
-
The molar ratio of chlorine to
product is 1 : 2 (Cl2 : ClF3)
-
Therefore the reacting gas volume
ratio is also 1 : 2
-
Therefore 50 cm3 of
chlorine will be converted to 50 x 2/1 =
100 cm3 of ClF3
-
-
-
(d) Calculate the volume of unreacted
fluorine.
-
mole ratio = volume ratio = 3 : 2
for F2 : ClF3 (selected reactant : product)
-
Since 100 cm3 of ClF3
was formed, from a ratio of 3 : 2 (3/2),
-
the volume of fluorine consumed
must be 3/2 x 100 = 150 cm3.
-
Therefore the unreacted fluorine
volume = the initial 200 - 150 =
50 cm3 of
excess unreacted fluorine.
-
Note: You can also
work it out from just the reactant gas volume ratio of 3 : 1 (F2
: Cl2 3/1)
-
The limiting reactant volume
of 50 cm3 of chlorine must therefore react with 50 x
3/1 = 150 cm3 F2
-
thus leaving 200 - 150 =
50 cm3 of
unreacted fluorine gas.
-
-
-
10.11
TOP OF PAGE
Self-assessment Quizzes
on REACTING GAS VOLUME RATIO
CALCULATIONS
type in answer
QUIZ or
multiple choice
QUIZ
See also for gas calculations
Advanced
notes on gas law calculations, kinetic
model theory of an IDEAL GAS & non-ideal gases
Moles and the molar volume of a gas, Avogadro's Law

Above is typical periodic table used in GCSE science-chemistry specifications in
doing reacting gas volume ratio calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
Revision notes on Avogadro's
Law, solving reacting gas volume ratio problems, how to calculate gas
volumes of products from reactant gas volume ratios, help when revising for AQA
A level &
GCSE chemistry, Edexcel A level & GCSE chemistry, OCR A level & GCSE gateway science chemistry,
OCR GCSE 21st century science chemistry notes for A level & GCSE 9-1 chemistry examination
practice questions on Gay-Lussac's Law, working out equations from reacting
gas volumes, using reactant product molar ratios to work out gas volumes,
relating molar ratios to gas volumes by interpreting the equation
Keywords: Quantitative chemistry
calculations Help for problem solving
in doing reacting gas volume calculations. Practice revision questions
on gas volume ratios of reactants and products from balanced equations,
using experiment data, making predictions. This page describes and
explains,
with fully worked out examples, how to calculate the volumes of gaseous
reactants or products
formed from given volumes of reactants or products. You need to know how
to apply Avogadro's Law and Gay-Lussac's Law of combining volumes. These
particular calculation methods simply use the ratio of reactant gases or
product gases given by the symbol equation. From reacting gas volumes
you can also use the mole concept to calculate the mass of products
formed. You can also work out the molecular formula of an unknown
gaseous compound from reacting volumes (historical method - much
more sophisticated methods these days!). Online practice exam chemistry
CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY
and basic starter chemical calculations for A level AS/A2/IB courses.
These revision notes and practice questions on how to do reacting gas
volume ratio chemical calculations (using Gay-Lussac's law) and worked
examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1)
chemistry science courses.
OTHER CALCULATION PAGES
-
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
-
Calculating relative
formula/molecular mass of a compound or element molecule
-
Law of Conservation of Mass and simple reacting mass calculations
-
Composition by percentage mass of elements
in a compound
-
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
-
Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination
-
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
-
Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
-
Moles and the molar volume of a gas, Avogadro's Law
-
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law
(this page)
-
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
-
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
-
Electrolysis products calculations (negative cathode and positive anode products)
-
Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
-
Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
-
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
-
Radioactivity & half-life calculations including
dating materials
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