REACTING GAS VOLUME RATIO
CALCULATIONS
GayLussac's Law
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Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level
Online Chemical Calculations
10.
Reacting gas volume ratios of reactants or products (Avogadro's Law,
GayLussac's Law)
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 GayLussac'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
Spotted any careless error?
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query?comment or request a type of
GCSE calculation not covered?
See also for gas calculations
Advanced
notes on gas law calculations, kinetic
model theory of an IDEAL GAS & nonideal gases
AND
Moles and the molar volume of a gas, Avogadro's Law
Selfassessment Quizzes: type in answer
Honly or
multiple choice
Honly

10.
Reacting gas volume ratios of reactants or products (Avogadro's Law,
GayLussac'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 lefthand gas syringe.

REACTING GAS VOLUMES and MOLE
RATIO

Historically GayLussac'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)}
+ NH_{3(g)} ==> NH_{4}Cl_{(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 GayLussac'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. 25cm^{3} + 25cm^{3}
==> solid product (no gas formed)

or 400dm^{3} + 400 dm^{3} ==>
solid product etc.

so, if 50 cm^{3} HCl
reacts, you can predict 50 cm^{3} of NH_{3} will react etc.
etc. !



Reacting gas
volume ratio calculation Example 10.2

Given the equation:
N_{2(g)}
+ 3H_{2(g)} ==> 2NH_{3(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. what volume
of hydrogen reacts with 50 cm^{3} nitrogen and what volume of
ammonia will be formed?

The ratio is 1 : 3 ==> 2, so you
multiply equation ratio numbers by 50 giving ...

50 cm^{3} nitrogen
+ 150 cm^{3} hydrogen (3 x 50) ==> 100 cm^{3} of
ammonia (2 x 50)



Reacting
gas volume ratio calculation Example
10.3

Given the equation: C_{3}H_{8(g)} + 5O_{2(g)}
==> 3CO_{2(g)} + 4H_{2}O_{(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 25cm^{3} of propane, C_{3}H_{8}.

Theoretical reactant
volume ratio is C_{3}H_{8 }: O_{2} is 1 : 5
for burning the fuel propane.

so actual ratio is 25
: 5x25, so 125cm^{3} oxygen is needed.



(b) What volume of carbon
dioxide is formed if 5dm^{3} of propane is burned?

Theoretical
reactantproduct volume ratio is C_{3}H_{8 }: CO_{2}
is 1 : 3

so actual ratio is 5
: 3x5, so 15dm^{3} carbon dioxide is formed.



(c) What volume of air (^{1}/_{5}th
oxygen) is required to burn propane at the rate of 2dm^{3} per minute
in a gas fire?

Theoretical reactant
volume ratio is C_{3}H_{8 }: O_{2} is 1 : 5

so actual ratio is 2
: 5x2, so 10dm^{3} oxygen per minute is needed,

therefore, since air is
only ^{1}/_{5}th O_{2}, 5 x 10 = 50dm^{3}
of air per minute is required.



Reacting
gas volume ratio calculation
Example 10.4

Given
the equation: 2H_{2(g)} + O_{2(g)} ==> 2H_{2}O_{(l)}

If 40 dm^{3} of
hydrogen, (at 25^{o}C 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 dm^{3} 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 gasvapour.

The theoretical gas volume ratio
of reactant hydrogen to product water is 1 : 1

Therefore, prior to condensation
at room temperature and pressure, 40 dm^{3} of water vapour is
formed.

1 mole of gas occupies 24 dm^{3},
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 =
30g of H_{2}O



Reacting
gas volume ratio calculation
Example 10.5

It was
found that exactly 10 cm^{3} of bromine vapour (Br_{2(g)})
combined with exactly 30 cm^{3} chlorine gas (Cl_{2(g)})
to form a brominechlorine compound BrCl_{x}.

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 BrCl_{x}

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 molecules) of the brominechlorine
compound being formed.

Reacting
gas volume ratio calculation
Example 10.6

Reacting
gas volume ratio calculation Example 10.7
Selfassessment Quizzes
[rgv] type in answer
Honly or
multiple choice
Honly
See also for gas calculations
Advanced
notes on gas law calculations, kinetic
model theory of an IDEAL GAS & nonideal gases
Moles and the molar volume of a gas, Avogadro's Law
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Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)

Reacting mass ratio calculations of reactants and products
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(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
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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

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ratios, Avogadro's Law
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