1
STUDENT NAME: ……………………………………………………………………….
CLASS: …………………………………
GRADE 9 – CHEMISTRY - UNIT 1- PRINCIPLES OF CHEMISTRY
CHAPTER 6 - CHEMICAL FORMULAE, EQUATIONS AND CALCULATIONS: PART 2 (PAGE 64 -67)
TUITORIAL – 03 - CALCULATIONS INVOLVING GAS VOLUMES
Avogadro's law:

The Avogadro's law says that …………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………..
 That is, 100 cm3 of hydrogen at some temperature and pressure contains exactly the same number of
molecules as there are in 100 cm3 of Cl2, or any other gas under those conditions, irrespective of the size
of the molecules.
e.g. Methane (CH4) reacts with oxygen to give carbon dioxide and water.
CH4(g)
1
+
:
2O2(g)
2
:
CO2(g)
1
+
:
2H2O(l)
2
 According to this equation, 100 cm3 of methane will react exactly with 200 cm3 of oxygen gas and 100
cm3of carbon dioxide gas will be formed.
CH4(g)
+
2O2(g)
CO2(g)
+
2H2O(l)
1
100 cm3
:
:
2
200 cm3
:
:
1
:
………………. :
2
…..............
UNITS OF VOLUME
 Volumes (of gases or liquids) are measured in
1 litre = 1 dm3 = 1000 cm3
cubic centimetres (cm 3)
or
cubic decimetres (dm3)
or
litres
(l)
2
Conversion between cm3 and dm3
1 dm3 = 1000cm3
1 mole of any gas occupies 24 dm3 = 24 000cm3 at room temperature and pressure (rtp)
 Room temperature and pressure (rtp) is usually taken as 20-25⁰C and 1 atmosphere pressure (atm).
END OF CHAPTER QUEATIONS (page 72)
THE VOLUME OCCUPIED BY 1 MOLE OF A GAS
 1 mole of any gas contains the same number of molecules and so occupies the same volume as 1 mole of
any other gas at the same temperature and pressure.
 At room temperature and pressure, the volume occupied by 1 mole of any gas is approximately
24 dm3 (24 000 cm3).
• The volume occupied by 1 mole of a gas is often called the molar volume.
The triangles shown below can be used to convert between volumes and numbers of moles.
CALCULATIONS WITH MOLAR VOLUME
CALCULATING THE VOLUME OF A CERTAIN NUMBER OF MOLES
Q1. Calculate the volume in dm3 of 0.20mol CO2 at rtp:
3
Volume = ………………………………………………. X ………………………………………………….
Since the volume in dm3 is needed, 24dm3 is used as the molar volume:
So, Volume =
…………… x …………………… = ………………………………………..
CALCULATING THE VOLUME OF A GIVEN MASS OF GAS
Q. Calculate the volume (in cm3) of 0.01 g of hydrogen at rtp (Ar: H = 1).
1mol H2 has a mass of 2g:
0.01 g of hydrogen is 0.01 mol = 0.005 mol
2
Since the volume in cm3 is needed, the molar volume is considered as ………………………………………..
Volume = …………………………………….. x …………………………………………….
=…………………………………….. x ……………………………………………..
So, 0.005mol of hydrogen occupies =…………………………………………
CALCULATING THE NUMBER OF MOLES FROM A VOLUME
Q. Calculate the amount of moles in 120 cm3 of carbon dioxide.
Since the volume is given in cm3 , 24000 cm3 should be used as the molar volume:
number of moles = volume of gas / molar volume
number of moles = …………………………………………………………=……………………………………………….
USING THE MOLAR VOLUME IN CALCULATIONS WITH CHEMICAL EQUATIONS
In order to do these calculations,
 Calculate the number of moles of anything which is possible.
 Use the chemical equation to deduce the number of moles of what is possible.
 Convert the number of moles to the required quantity, e.g. a mass or a volume.
Q. Calculate the volume of carbon dioxide produced at room temperature and pressure when an excess
of dilute hydrochloric acid is added to 1.00g of calcium carbonate. (Ar: C = 12, 0 = 16, Ca = 40; molar
volume = 24 dm3 at rtp.)
The equation for the reaction is,
Molar ratio
CaCO3(s) + 2HCl(aq)
1
:
2
CaCl2(aq) + CO2(g) + H2O(1)
:
1
:
1
:
1
4
The mass of calcium carbonate (CaCO3) is given. So, the number of moles of calcium carbonate can be
calculated:
• According to the equation, 1 mol of CaCO3 produces 1 mol of CO2, so, 0.0100 mol of CaCO3 will produce
the same number of moles of CO2, that is, 0.0100mol.
CaCO3(s) + 2HCl(aq)
1
:
2
:
0.0100 mol
;
Molar ratio
CaCl2(aq) + CO2(g) + H2O(1)
1
:
1
: 1
0.0100 mol
So, the volume occupied by 0.0100 mol of CO2 is,
volume of gas
=
number of moles x molar volume
= …………………. X …………………. = ……………………………dm3
A PROBLEM INVOLVING MAKING HYDROGEN
• Aluminium reacts with dilute hydrochloric acid according to the equation.
2Al(s)
+
2
:
6HCl(aq)
6
:
2AICl3(aq)
+
3H2(g)
2
:
3
Q. What mass of aluminium would you need to add to an excess of dilute hydrochloric acid so that you
produced 100 cm3 of hydrogen at room temperature and pressure? (Ar of Al = 27; molar volume = 24000
cm3 at rtp.)
• Calculate the number of moles of the hydrogen gas.
The volume of hydrogen gas is given in cm3. Use the molar volume of a gas as 24 000 cm3:
Number of moles of H2 gas = Volume of gas
=……………………………………………………..
Molar volume
We could also write this number in standard form as ………………………………………………………………...
• The number of moles of Al will be 2/3 times the number of moles of H2 according to the above equation.
2Al(s)
+
2
:
? mol
6HCl(aq)
6
:
2AICl3(aq)
+
3H2(g)
2
:
3
……………………………..mol
• The number of moles of aluminium required to produce 0.00417 mol of H2 is ………………………………………..
……………………………………………………………………………………………………………………………………………………………………
Then, Mass of Aluminium = number of moles x mass of 1 mol = ………………………………… = ………………………
THE END