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 9.
The molar gas volume in calculations, Avogadro's Law
Avogadro's Law states that equal volumes of gases under the same conditions of temperature
and pressure contain the same number of molecules. So the volumes have equal moles of
separate particles in them. One mole of any gas (or the formula mass in g), at the same temperature and pressure occupies the
same volume . This is 24dm3 (24 litres) or 24000 cm3, at room temperature
(25oC/298K) and
pressure (101.3kPa/1atmosphere). The molar volume for s.t.p is 22.4 dm3
(22.4 litres) at 0oC and 1atmosphere pressure. Historically, s.t.p
unfortunately stands for standard temperature and pressure, but these days 25oC/298K
is usually considered the standard temperature.
-
Some handy relationships for
substance Z below:
-
moles Z = mass of Z
gas (g) / atomic or formula mass of gas Z (g/mol)
-
mass of Z in g =
moles of Z x atomic or formula mass
of Z
-
atomic or formula mass of Z
= mass of Z / moles of
Z
-
1 mole = formula
mass of Z
in g.
-
gas volume of
Z
= moles of Z
x molar volume
-
Note (i): In the following examples, assume you are
dealing with room temperature and pressure i.e. 25oC and 1 atmosphere
pressure.
-
Note (ii):
-
Apart from solving the problems using the
mole concept (method (a) below, and reading any equations
involved in a 'molar way' ...
-
It is also possible to solve them without using the mole
concept (method (b) below). You still use the molar volume itself,
but you think of it as the volume occupied by the formula mass of the
gas in g and never think about moles!
-
Example 9.1: What is the volume of 3.5g of hydrogen? [Ar(H)
= 1]
-
Example 9.2: Given the equation MgCO3(s) + H2SO4(aq) ==>
MgSO4(aq)
+ H2O(l) +CO2(g)
-
What mass of magnesium carbonate is needed to make 6 dm3
of carbon dioxide? [Ar's: Mg = 24, C = 12, O = 16, H =1 and S
= 32]
-
method (a):
-
since 1 mole = 24 dm3, 6 dm3 is
equal to 6/24 = 0.25 mol of gas
-
From the equation, 1 mole of MgCO3 produces 1
mole of CO2, which occupies a volume of 24 dm3.
-
so 0.25 moles of MgCO3 is need to make 0.25
mol of CO2
-
formula mass of MgCO3 = 24 + 12 + 3x16 = 84,
-
so required mass of MgCO3 = mol x formula
mass = 0.25 x 84 = 21g
-
method (b):
-
converting the equation into the required reacting masses ..
-
formula masses: MgCO3 = 84 (from above), CO2 = 12 +
2x16 = 44
-
MgCO3 : CO2 equation ratio is 1 :
1
-
so 84g of MgCO3 will form 44g of CO2
-
44g of CO2 will occupy 24dm3
-
so scaling down, 6 dm3 of CO2 will
have a mass of 44 x 8/24 = 11g
-
if 84g MgCO3 ==> 44g of CO2,
then ...
-
21g MgCO3
==> 11g of CO2 by solving the ratio, scaling down by
factor of 4
-
Example 9.3: 6g of a hydrocarbon gas had a volume of 4.8
dm3. Calculate its molecular mass.
-
method (a):
-
1 mole = 24 dm3, so moles of gas = 4.8/24 = 0.2 mol
-
molecular mass = mass in g / moles of gas
-
Mr = 6 /
0.2 = 30
-
i.e. if 6g = 0.2 mol, 1 mol must be
equal to 30g by scaling up
-
method (b):
-
6g occupies a volume of 4.8 dm3
-
the formula mass in g occupies 24 dm3
-
so scaling up the 6g in 4.8 dm3
-
there will be 6 x 24/4.8 = 30g in 24 dm3
-
so the
molecular or formula
mass = 30
-
Example 9.4: Given the equation ... (and Ar's
Ca = 40, H = 1, Cl = 35.5)
-
Ca(s) + 2HCl(aq) ==> CaCl2(aq)
+ H2(g)
-
What volume of hydrogen is formed when (i) 2g of calcium
is dissolved in excess hydrochloric acid?, (ii) 0.25 moles of hydrochloric
acid reacts with calcium?
-
(i) method (a):
-
3g Ca = 3/40 = 0.075 mol Ca
-
from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H2
-
so 0.075 mol Ca produces 0.075 mol H2
-
so volume H2
= 0.075 x 24 = 1.8 dm3
(or 1800 cm3)
-
(i) method (b):
-
from equation 1 Ca ==> 1 H2 means 40g
==> 2g
-
so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H2
-
2g H2 has a volume 24 dm3, so
scaling down ...
-
0.15g H2 has a
volume of (0.15/2) x 24 = 1.8 dm3
(or 1800 cm3)
-
(ii) method (a) only:
-
from equation: 2 moles HCl ==> 1 mole H2
(mole ratio 2:1)
-
so 0.25 mol HCl ==> 0.125 mol H2, volume 1
mole gas = 24 dm3
-
so volume H2
= 0.125 x 24 = 3 dm3
-
Example 9.5: Given the equation ... (and Ar's
Mg = 24, H = 1, Cl = 35.5)
Example 9.6: A small teaspoon of sodium hydrogencarbonate
(baking
soda) weighs 4.2g. Calculate the moles, mass and volume of carbon dioxide
formed when it is thermally decomposed in the oven. Assume room
temperature for the purpose of the calculation.
-
2NaHCO3(s) ==> Na2CO3(s)
+ H2O(g) + CO2(g)
-
Formula mass of NaHCO3 is 23+1+12+(3x16) = 84 =
84g/mole
-
Formula mass of CO2 = 12+(2x16) = 44 = 44g/mole
(not needed by this method)
-
In the equation 2 moles of NaHCO3 give 1 mole
of CO2 (2:1 mole ratio in equation)
-
Moles NaHCO3 = 4.2/84 = 0.05 moles ==>
0.05/2 = 0.025 mol CO2 on decomposition.
-
Mass = moles x formula mass, so mass CO2 =
0.025 x 44 = 1.1g CO2
-
Volume = moles x molar volume = 0.025 x 24000 = 600 cm3
of CO2
-
See also QA7.1 on the mole
introduction page
Self-assessment Quizzes
[mvg] type in answer
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or
multiple choice
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