Moles and reactions
At the end of this section, you should
be able to deduce the quantities of
reactants and products from balanced
equations
Key information
Stoichiometry is the molar relationship between the relative
quantities of substances taking part in a reaction.
Balanced equation gives information about:
* the reacting quantities that are needed to prepare a required
quantity of a product
* the quantity of products formed by reacting together of known
quantities of reactants.
In dealing with reacting quantities, the following steps should be
considered
1. Write the balanced reaction equation if not given
2. Work out the moles of the known reacting quantities
3. Work out the mass or any other quantity required
Recall
There are three ways to work out an amount in moles
1. From mass:
n=m
or
M
m=nxM
2. From gas volumes:
n = V(dm3) or V(dm3) = n x 24.0
24.0
3. From solutions:
n = c x V(dm3)
Examples (reacting masses)
1.
8.00g of aluminium reacted with excess oxygen to
produce aluminium oxide.
a) write a balanced equation
b) find the number of moles of aluminium oxide formed.
c) what was the mass of aluminium oxide produced?
Answer
a) 4Al
+
4mol
2 mol
b)
3O2  2Al2O3
3 mol
1.5 mol
2 mol
1 mol
Moles of aluminium, n = m
M
n = 8.00 = 0.2963
27
From the reaction, 2 mols of Al produces 1 mol of Al2O3
Therefore 0.2963 mols of Al will produce = 0.2963
.
2
= 0.1482mols Al2O3.
c) mass = n x M
M(Al2O3) = 27 x 2 + 16 x3 = 102
mass of Al2O3 produced = 102 x 0.1482
= 15.12g
2. What mass of copper oxide is obtained by heating 4.94g of
copper carbonate?
CuCO3(s)  CuO(s) + CO2(g)
( hint. Find the moles, mole ratio of quantities and the mass)
Answer
from n = m ÷ M
M(CuCO3) =64 + 12 + 16 x 3 = 124
n = 4.94 ÷ 124
= 0.0398 mol of CuCO3
From the reaction, 1 mol of CuCO3 produces 1 mol of CuO
Therefore 0.0398 mol of CuCO3 produces 0.0398 mol of CuO
mass of CuO = n x M
M(CuO) = 64 + 16 = 80
= 0.0398 x 80
= 3.18g of CuO
• 3. Sodium and chorine react to form sodium chloride:
2Na (s)+ Cl2(g)  2NaCl(s)
(a) Calculate the amount, in mol, in 2.925 g NaCl.
n = m/M =
2.925 = 2.925 = 0.0500 mol
23.0 + 35.5 58.5
(b) Calculate the amounts, in mol, of Na and Cl2 that would form 2.925 g NaCl.
equation:
2Na (s)+ Cl2(g)  2NaCl(s)
amounts in equation 2 mol : 1 mol : 2 mol
Amounts required :
0.050mol 0.025mol 0.050mol
(c) Calculate the masses of Na and Cl2 that would form 2.925 g NaCl.
Na : mass = n x M = 0.050 x 23 = 1.15 g
Cl2 : mass = n x M = 0.025 x (35.5+35.5) = 0.025 + 71 = 1.775 g
(4) Zinc carbonate decomposes to produces carbon
dioxide and 4.2 g zinc oxide.
(a) write a balanced reaction equation
(b) find the moles of zinc carbonate that
decomposed
(c) what mass of zinc carbonate produced the
4.20 g of zinc oxide.
Answers
(a) Equation: ZnCO3  ZnO + CO2
(b) n = m
M
M(ZnO) = 65.4 + 16.0 = 81.4
mole of ZnO, n = 4.20 = 0.0516 mol
81.4
reaction equation:
ZnCO3  ZnO + CO2
amounts in equation:
1 mol:
1 mol: 1 mol
amounts required:
0.0516
0.0516 0.0516
therefore amount of zinc carbonate that decomposed was 0.0516
mol
(c)mass = n x M
M(ZnCO3) = 65.4 + 12.0 + (16.0x3) = 125.4
mass of zinc carbonate = 125.4 x 0.0516 = 6.47 g
therefore 6.47 g zinc carbonate is required
Reacting masses and gas volumes
If a substance is a solid, mass is used to measure the
amount needed for a reaction. However, if a reactant or
product is a gas, it unlikely it could be weighed; it would
be easier to measure its volume.
According to Avogadro’s Law: Equal volumes of gases
measured at the same temperature and pressure contain
the same number of molecules.
This is extremely useful when relating the volume of gas
to an equation.
Example
What volume of oxygen reacts exactly with 100cm3 of
hydrogen at the same temperature and pressure?
Answer
the equation for the reaction is:
2H2(g) + O2(g)  2H2O(l)
the mole ratio
2 mol: 1 mol: 2 mol
volume required
100cm3
50cm3
100cm3
The equation indicates that half the number of oxygen
molecules as hydrogen molecules is needed. Therefore,
50cm3 of oxygen reacts with 100cm3 of hydrogen
2. What volume of oxygen reacts exactly with 30 cm3 of methane and what
volume of carbon dioxide is obtained at the same temperature and pressure?
Answer
The equation for the reaction is:
CH4 (g) + 2O2 (g)  CO2(g) + 2H2O(l)
the mole ratio is
1 mol:
2 mol:
1 mol:
2 mol
volume of gas
30cm3
60cm3
30cm3
60cm3
therefore 60 cm3 of oxygen reacted with the 30 cm3 of methane to produce 30 cm3
of carbon dioxide
Molar volume
If a volume of gas contains a fixed number of molecules, then it is
useful to know what volume contains the Avogadro number of
molecules. This volume is known as Molar volume. Molar volume
depends on standard temperature of 0oC (273K) and a pressure of
101.3 kPa.
Under the above conditions, molar volume is taken as 24000 cm3 or
(24 dm3)mol-1.
Main statement
At room temperature and pressure ( r.t.p), 1 mol of a gas occupies
24000 cm3 or 24dm3
(Hint: you need to find the number of moles of the gas in order to calculate the molar
volume at r.p.t, and where necessary use mole ratio from the balanced reaction
equation)
EXAMPLES
1. 2.50g of calcium carbonate was heated and decomposed as in the
equation below.
CaCO3(s)  CaO(s) + CO2(g)
(a) calculate the amount, in mol, of CaCO3 that decomposed.
n= m =
M
2.50
= 2.50 = 0.025 mol
{40 + 12 + (3x16)}
100
(b) calculate the amount, in mol, of carbon dioxide formed.
Reaction equation:
CaCO3(s)  CaO(s) + CO2(g)
mole ratio:
1 mol:
1 mol: 1 mol
amounts of mole required
0.025
0.025
0.025
i.e. 0.025 mol of carbon dioxide formed.
(c) calculate the volume of carbon dioxide that formed.
V = n x 24.0dm3 = 0.025 x 24 = 0.60 dm3
2. Calculate the volume of hydrogen produced when excess dilute
sulphuric acid is added to 5.00 g of zinc. Assume that under the
conditions of the reaction 1 mol of gas has a volume of 24 dm3.
Answer:
amount in moles of zinc used, n = m = 5.00 = 0.0765 mol
M
65.4
the equation is:
Zn(s) + H2SO4(aq)  ZnSO4(aq) + H2(g)
mole ratio
1 mol 1 mol
1 mol
1 mol
amounts required
0.0765 mol
0.0765 mol
it follows that 0.0765 mol of hydrogen is produced
molar volume = 24.0 dm3
volume of hydrogen produced, V = n x 24.0 = 0.0765 x 24
= 1.836dm3
3. Calculate the volume of hydrogen produced when excess dilute sulphuric acid is
added to 5.00 g of aluminium. Under the conditions of the reaction, 1 mol of gas has
a volume of 24.0 dm3.
Answer
amount of mole of Al used, n = m = 5.00 = 0.185 mol
M 27.0
the equation for the reaction is:
2Al(s) + 3H2SO4(aq)  Al2(SO4)3(aq) + 3H2(g)
mole ratio
2 mol 3 mol
1 mol
3 mol
simplest ratio
1 mol 1.5 mol
0.5 mol
1.5mol
the equation indicates that 1 mol of aluminium produces 1.5 mol of hydrogen.
Amount in moles of hydrogen produced = 1.5 x 0.185 = 0.278 mol
hence, volume of hydrogen produced, V = n x 24.0 = 0.278 x 24 = 6.667 dm3
4. When reacted with excess dilute hydrochloric acid, what mass of
calcium carbonate produces 1 dm3 of carbon dioxide? Assume that 1
mol of of gas has a volume of 24 dm3
Answer
amount in moles of carbon dioxide = 1 = 0.0417 mol
24
the equation for the reaction is:
CaCO3(s) + 2HCl(aq)  2CaCl(aq) + CO2(g) + H2O(l)
mole ratio: 1 mol
2 mol
2 mol
1 mol
1 mol
from the reaction equation 1 mol of carbon dioxide is produced by the
reaction of 1 mol of calcium carbonate with excess hydrochloric acid.
Therefore amount in moles of CaCO3 used = 0.0417 mol
mass of CaCO3 used, m = n x M
M(CaCO3) = 40. 0 + 12.0 + (3 x 16.0) = 100
hence, mass of CaCO3 used = n x M = 0.0417 x 100 = 4.17 g
5. A gas syring has a maximum capacity of 100 cm3. Calculate the
mass of copper(II) carbonate that would have to be heated to produce
enough carbon dioxide to just fill the syringe at r.t.p
Answer
syringe volume of 100 cm3 if equivalent to 100 = 0.00417 mol
24000
thus, volume of carbon dioxide to fill the syringe is 0.00417 mol
the equation for the reaction is:
CuCO3(s)  CuO(s) + CO2(g)
mole ratio
1 mol
1 mol
1 mol
amounts required
0.00417 mol
0.00417 mol
thus, 0.00417mol of carbon dioxide is obtained by heating 0.00417mol
of copper carbonate.
Mass of copper carbonate, m = n x M
= 0.00417 x ( 63.5 + 12 + 16x3)
= 0.00417 x 123.5
= 0.515 g
6. Calculate the volume of 8 g of methane at r.t.p
Answer
amount in mol of methane, CH4, n = m =
8
M 12 + (4 x 1)
amount in mol of methane = 0.5 mol
volume of 1 mol of methane at r.t.p = 24.0 dm3
hence, volume of 0.5 mol of methane = 0.5 x 24
= 12 dm3
=8
16
Reacting mass, gas and solution volumes
You will need to recall some of the formulae used
earlier on and use them in such calculations.
Rearrangement of the formulae may be necessary in
some instances in order to do other calculations.
Recall
1. From mass:
n=m
or
M
m=nxM
2. From gas volumes:
n = V(dm3) or V(dm3) = n x 24.0
24.0
3. From solutions:
n = c x V(dm3) or c =
n
V(dm3)
1.
A chemist reacted 0.23g of sodium with water to form 250 cm3 of aqueous
sodium hydroxide. Hydrogen gas was also produced. The equation is shown
below. 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
(a) calculate the amount, in mol, of Na that reacted.
n = m = 0.23 = 0.010 mol
M 23
(b) Calculate the volume of H2 formed at room temperature and pressure.
The equation for the reaction is 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
mole ratio
2 mol
1 mol
amounts required
0.010mol
0.005mol
thus, 0.010 mol of Na produces 0.005 mol H2 gas.
Volume of hydrogen gas formed at r.t.p, V = n x 24000 cm3
= 0.005 x 24000 = 120 cm3
(c ) Calculate the concentration, in mol/dm3, of NaOH(aq) formed.
Reaction equation:
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
mole ration:
2 mol
2 mol
amount required
0.010 mol
0.010mol
thus, 0.010 mol Na formed 0.010 mol of NaOH solution
concentration of NaOH, in mol/dm3, c = n
= 0.010 mol = 0.04 mol/dm3
V dm3 0.250 dm3
2. Balance the equation below.
MgCO3(s) + HNO3(aq)  Mg(NO3)2(aq) + H2O(l) + CO2(g)
(b) 2.529 g of MgCO3 reacts with an excess of HNO3.
(i) what volume of CO2, measured at RTP, is formed?
(ii) The final volume of the solution is 50 cm3. What is the
concentration, in mol/dm3, of Mg(NO3)2 formed
Now try questions 1 – 3
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