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. Q: what volume
of hydrogen reacts with 50 cm^{3} 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 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 (sometimes you need to
think 'outside the box' in some context e.g.

Some industrial processes produce toxic
carbon monoxide as a byproduct 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) + O_{2}(g)
===> 2CO_{2}(g)

(a) If carbon monoxide is produced in an
industrial process at the rate of 50 dm^{3} 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 O_{2} to form 2 mol of CO_{2}

Therefore 2 volumes of CO reacts with 1
volume of O_{2} to form 2 volumes of CO_{2} by ratio.

Therefore volume of O_{2} needed
= half the volume of CO produced.

Therefore for every 50 dm^{3} of
CO produced you need 25 dm^{3} of oxygen to burn it.

Therefore you need to pump in oxygen at
the rate of 25 dm^{3}/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 dm^{3} air/min



Reacting
gas volume ratio calculation Example
10.4

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 15 dm^{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 =
50 dm^{3}
of air per minute is required.



Reacting
gas volume ratio calculation
Example 10.5

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 =
30 g of H_{2}O



Reacting
gas volume ratio calculation
Example 10.6

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 two molecules) of the brominechlorine
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 cm^{3} of hydrogen and 20 cm^{3}
of carbon monoxide?

(i) hydrogen: H_{2}(g) + ½O_{2}(g)
===> H_{2}O(l)

mole ratio/gas volume ratio H_{2}
: O_{2} is 1 : 0.5

therefore reacting volume ratio must be
the same for 10 cm^{3} : y cm^{3}

so y = 10 x 0.5 / 1 =
5 cm^{3}
(10/2)



(ii) carbon monoxide: CO(g) + ½O_{2}(g)
===> CO_{2}(g)

mole ratio/gas volume ratio CO : O_{2}
is 1 : 0.5

therefore reacting volume ratio must be
the same for 20 cm^{3} : z cm^{3}

so z = 20 x 0.5 / 1 =
10 cm^{3}
(20/2)



(iii) Therefore total volume oxygen = y +
z = 5 + 10 = 15 cm^{3} O_{2}

(iv) Assuming air is 1/5th (20%) oxygen,
the volume of air required would be 5 x 15 = 75 cm^{3}
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 dm^{3} of methane and 10 dm^{3}
of propane?

(i) methane: CH_{4(g)} + 2O_{2(g)}
==> CO_{2(g)} + 2H_{2}O_{(l)}

mole ratio = gas volume ratio CH_{4}
: O_{2} is 1 : 2

therefore reacting volume ratio is 20
: y

so y = 20 x 2 / 1 =
40 dm^{3}

(ii)
propane: C_{3}H_{8(g)} + 5O_{2(g)}
==> 3CO_{2(g)} + 4H_{2}O_{(l)}

mole ratio = gas volume ratio C_{3}H_{8}
: O_{2} is 1 : 5

therefore reacting volume ratio is 10
: z

so z = 10 x 5 / 1 =
50 dm^{3}

(iii) Therefore total volume oxygen = y +
z = 40 + 50 = 90 dm^{3} O_{2}

(iv) Assuming air is 1/5th (20%) oxygen,
the volume of air required would be 5 x 90 =
450 dm^{3}
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: 3F_{2}(g)
+ Cl_{2}(g) ===> 2ClF_{3}(g)

If 200 cm^{3} of fluorine gas is
reacted with 50 cm^{3} 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 (Cl_{2} : ClF_{3})

Therefore the reacting gas volume
ratio is also 1 : 2

Therefore 50 cm^{3} of
chlorine will be converted to 50 x ^{2}/_{1} =
100 cm^{3} of ClF_{3}



(d) Calculate the volume of unreacted
fluorine.

mole ratio = volume ratio = 3 : 2
for F_{2} : ClF_{3} (selected reactant : product)

Since 100 cm^{3} of ClF_{3}
was formed, from a ratio of 3 : 2 (^{3}/_{2}),

the volume of fluorine consumed
must be ^{3}/_{2} x 100 = 150 cm^{3}.

Therefore the unreacted fluorine
volume = the initial 200  150 =
50 cm^{3} of
excess unreacted fluorine.

Note: You can also
work it out from just the reactant gas volume ratio of 3 : 1 (F_{2}
: Cl_{2} ^{3}/_{1})

The limiting reactant volume
of 50 cm^{3} of chlorine must therefore react with 50 x
^{3}/_{1} = 150 cm^{3} F_{2}

thus leaving 200  150 =
50 cm^{3} of
unreacted fluorine gas.



10.11
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reactants or products
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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
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gaseous compound from reacting volumes (historical method  much
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