Reacting gas volume ratios of reactants or products (Avogadro'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
Historically Gay-Lussac's Law of volumes states that 'gases combine with each other in
simple proportions by volume'.
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
of the equation or the relative moles of reactants and products
automatically gives us the gas volumes ratio
of reactants and products,
if all the gas volumes are measured at the same temperature and pressure.
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
or (ii) a mole ratio,
which applies to anything in the equation, whether (g), (l) or (s).
volume ratio calculation Example 10.1
Given the equation:
+ NH3(g) ==> NH4Cl(s)
hydrogen chloride gas combines with 1 mole of ammonia gas to give 1 mole
of ammonium chloride solid.
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.
The note the
volume ratio calculation Example 10.2
Given the equation:
+ 3H2(g) ==> 2NH3(g)
1 mole of
nitrogen gas combines with 3 mols of hydrogen gas to form 2 mol of a
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 cm3 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 cm3 nitrogen
+ 150 cm3 hydrogen (3 x 50) ==> 100 cm3 of
ammonia (2 x 50)
gas volume ratio calculation Example
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
(a) What volume of oxygen
is required to burn 25cm3 of propane, C3H8.
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?
reactant-product volume ratio is C3H8 : CO2
is 1 : 3
so actual ratio is 5
: 3x5, so 15dm3 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?
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 = 50dm3
of air per minute is required
gas volume ratio calculation Example 10.4
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
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
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 = moles x formula mass
mass water formed = 1.666 x 18 =
30g of H2O
gas volume ratio calculation Example 10.5
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 molecules) of the bromine-chlorine
compound being formed.
gas volume ratio calculation Example 10.6
gas volume ratio calculation Example 10.7
[rgv] type in answer
OTHER CALCULATION PAGES
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
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
moles) and brief mention of actual percent % yield and theoretical yield,
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
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
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
do volumetric titration calculations e.g. acid-alkali titrations
(and diagrams of apparatus)
Electrolysis products calculations (negative cathode and positive anode products)
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
Energy transfers in physical/chemical changes,
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity & half-life calculations including
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