CHAPTER 10
Suppose you are given the equation below,
2NO(g) + O2(g) uv
2NO2(g)
The above equation is interpreted as 2 molecules of NO (nitrogen
monoxide) react with 1 molecule of O2 (oxygen) to produce 2 molecules of
NO2 (nitrogen dioxide). These coefficients indicate relative number of
reactant and product molecules.
Interpretation of chemical equation coefficients
For the General equation
2A + 3B
C + 2D
The ratio of molecules is
2 : 3
:
1 : 2
The ratio of moles is
2: 3
:
1 : 2
The ratio of volumes of gas is
2: 3
:
1 :2
The ratio of Avogadro’s number is
2: 3
:
1 : 2
VERIFYING THE CONSERVATION OF MASS LAW
Law of conservation of mass states that mass is neither created nor
destroyed during a chemical reaction. The combined mass of reactants must
be equal to the combined mass of products.
Example, 2 NO(g) + O2(g)
2NO2(g)
2(30.01g) + 1(32.00g)
2(46.01g)
92.02g
92.02g
MOLE-MOLE RELATIONSHIPS
Considering the reaction below:
N2(g) + O2(g)
2NO(g)
1 mol of N2 reacts with 1 mol of O2, the corresponding mole ratios are:
1 mol N2 = 1 mol O2
1 mol O2
1 mol N2
1 mol of N2 produces 2 mol of NO, the corresponding mole ratios are:
1 mol N2
= 2 mol NO
2 mol NO
1 mol N2
1 mol of O2 produces 2 mol of NO, the corresponding mole ratios are:
1 mol O2
= 2 mol NO
2 mol NO
1 mol O2
TYPES OF STOICHIOMETRY PROBLEMS.
The term stoichiometry is used to refer to the relationship between
quantities in a chemical reaction according to the balanced chemical
equation. The word is coined from Greek words meaning “element”
“measure”.
Stoichiometry problems can be classified as one of these basic types:massmass, mass- volume, volume- volume problems.
MASS- MASS PROBLEMS.
In this type of calculation, the unknown mass of substance is calculated
from given mass of a reactant or product in a chemical equation as follows:
 Convert the given mass of substance to moles using the molar mass of
the substance as a unit factor.
 Convert the moles of the given to moles of the unknown using the
coefficients in the balanced equation.
 Convert the moles of the unknown to gram using the molar mass of
the substance as a unit factor.
Example: calculate the mass of Aluminium, considering the high
temperature conversion of 14.4g of iron (II) oxide to elemental iron with
aluminium metal. Given equation
3Fe(l) + 2Al(l)
3Fe(l) + Al2O3(l)
MM FeO = 71.85g/mol
14.4g FeO x 1 mol FeO = 0.200 mol FeO
71.85g FeO
From the equation, 3 mol of FeO = 2 mol of Al
0.200 mol of FeO x 2 mol Al
= 0.134 mol Al
3 mol of FeO
MM of Al = 28.98g/mol (using the unit factor)
0.134mol Al x 28.98g Al
= 3.60g Al
1 mol Al
14.4g of FeO reacts with 3.60g of Al.
MASS- VOLUME PROBLEMS.
This involves an unknown volume of substance is calculated from given
mass of reactant or product in a chemical equation.
 Convert the given mass of substance to moles using the molar mass of
the substance as a unit factor.
 Convert the moles of the given to moles of the unknown using
coefficients in the balanced equation.
 Convert the moles of the unknown to liters using the molar volume of
a gas as a unit factor. At standard temperature and pressure (STP), the
molar volume of a gas is 22.4 l/mol.
Example, Consider the reaction of 0.165g of aluminium metal with dilute
hydrochloric acid, the balanced equation,
2Al(s) + 6HCl(aq)
2AlCl3(aq) + 3H2(g)
Calculate the volume of hydrogen gas produced by the reaction at STP.
Number of mol of Al (mm of Al = 26.98g/mol)
0.165g Al x 1 mol Al
= 0.00611mol Al
26.98g Al
Number of moles H2, using the coefficients in balanced equation,
0.00611 mol Al x 3 mol H2
= 0.00917 mol H2
2 mol Al
Volume of H2, applying molar volume, 22.4l/1 mol,
0.00917 mol H2 x 22.4L
= 0.250L of H2
1 mol H2
VOLUME- VOLUME PROBLEMS
In a volume- volume problem, we take advantage of Gay- Lussac’s
law of combining volumes which states that volumes of gases, under similar
conditions, combine in small whole number ratios.
H2(g) + Cl2(g)
2HCl(g)
10mL 10mL
20mL
1
: 1
:
2
2H2(g) + O2(g)
2H2O(g)
100mL 50mL
100mL
2
:
1
:
2
Since the volumes of gases combine in the same whole number ratio
as the coefficients in the balanced equation, we can solve this type of
problem in one conversion step. We simply multiply the given volume by
the ratio of the coefficients in the equation.
Example, calculate the liters of oxygen gas that react with 37.5 L of
sulfur dioxide, in the following reaction:
2SO2(g) + O2(g) Pt/
2SO3(g)
2L of SO2 = 1L of O2 ( from the equation)
37.5L of SO2 x 1L of O2
= 18.8L of O2
2L of SO2
If we need to calculate the volume of SO3 produced, 2L of SO2 = 2L of SO3
( from the equation)
37.5L of SO2 x 2L of SO3
= 37.5L of SO3
2L of SO2
LIMITING REACTANT CONCEPT
A limiting reactant is the reactant that controls the amount of products
formed. We need to identify the limiting reactant in a reaction in order to
correctly calculate the amount of product. Suppose we have a reaction,
Fe +
S
FeS
0.25mol 3.00mol
0.00mol
0.25mol 0.25mol
0.25mol
Since the reaction is ratio 1: 1 : 1, we can conclude that FeS will be
0.25mol. Thereby leaving sulphur with excess of (3.00 – 0.25 = 2.75 mol).
We can therefore conclude that Fe is the limiting reagent.
Example: A 5.00mol sample of iron(III) oxide is heated with 5.00mol
aluminium metal and converted to molten iron. Identify the limiting reactant,
and calculate the number of moles of iron produced, given the equation :
Fe2O3(l) + 2Al(l)
2Fe(l) (l) + Al2O3(l)
Calculating the number of moles of Fe,
5.00mol Fe2O3 X 2mol Fe
= 10 mol Fe
1 mol Fe2O3
5.00mol Al X 2mol Fe
= 5 mol Fe
2 mol Al
Therefore the limiting reactant is Al and the number of mole of Fe produced
was 5 mol.
When the mass is given, we determine the limiting reactant as follows:
I.
Calculating the mass of product that can be produced from the first
reactant
 Calculate the moles of reactant.
 Calculate the moles of product.
 Calculate the mass of product.
II.
Calculate the mass of product that can be produced from second
Reactant.
 Calculate the moles of reactant.
 Calculate the moles of product
 calculate the mass of product
III. State the limiting reactant and the corresponding mass of product
Formed. The limiting agent gives the least amount product. The
actual mass of product obtained from the chemical reaction is the lesser of
the two product masses calculated in steps I and II. Suppose we wish to find
how much molten iron is produced from the reaction of 25.0g of FeO with
25.0g of Al.
FeO(l) + 2Al(l)
3Fe(l) + Al2O3(l)
g FeO to mol FeO to mol Fe to g Fe
Steps:
Ia
Ib
Ic
Mm of Fe = 55.85g/mol and Mm of FeO = 71.85g/mol
25.0g FeO x 1 mol FeO = 0.348 mol FeO
71.85g FeO
Calculating number of moles of Fe, (relate mole ratios in the equation)
0.348mol FeO x 3mol Fe = 0.348 mol Fe
3mole FeO
Calculating the mass of Fe,
0.348mol Fe x 55.85g Fe
= 19.4g Fe
1 mol Fe
g Al to mol Al to mol Fe to g Fe
Steps:
IIa
IIb
IIIb
Mm of Al = 27.98g/mol and Mm of Fe = 555.85g/mol
25.0g Al x 1 mol Al
= 0.893 mol Al
27.98g Al
Calculating number of moles of Fe,(relate mole ratios in the equation)
0.893 mol Al x 3 mole Fe
= 1.340 mol Fe
2 mole Al
Calculating the mass of Fe,
1.340 mole Fe x 55.85g Fe
= 74.9 g Fe
1 mol Fe
FeO is the limiting reactant while mass of Fe yielded was 19.4g.
Limiting Reactant Involving Volumes of Gas
We assume the temperature and pressure remain constant and use the
following approach to determine the limiting reactant.
I.
Calculating the volume of product that can be produced from first
reactant. Recall the coefficient in the balanced equation is in the
same ratio as the volume of gases.
II.
Calculate the volume of product that can be produced from the
second reactant. Again, the coefficients in the balanced equation
are in the same mole ratio as the volumes of gases.
III. State the limiting reactant and the corresponding volume of
product. The limiting reactant is the gas that gives the least amount
of product. The actual volume of product obtained from the
chemical reaction is the lesser of two volumes calculated in steps I
and II.
Example: Suppose 5.00L of NO gas reacts with 5.00L of oxygen gas,
which is the limiting reactant and what is the volume of product
produced? 2NO(g) (g) + O2(g)
2NO2(g)
Step I: 5.00L NO x 2L NO2
= 5.00L NO2
2L NO
Step II: 5.00L O2 x 2L NO2
= 10.00 L NO2
1L O2
Step III: The limiting reactant is NO and the volume of NO2 produced was
5.00L.
PERCENTAGE YIELD
Percentage yield = actual yield
x 100
Theoretical yield
The theoretical yield is the calculated yield of the product while the actual
yield is the result you obtained in the lab by going through the procedures
required for the product formation. In an experiment, some errors lead to
high results while other errors lead to low results. High results will be
obtained when impurities are trapped or when the precipitates are not
completely dry.
Example: 15.0kg of ammonia gives an actual yield of 65.3kg of
ammonium nitrate, the calculated yield of ammonium nitrate for experiment
is 70.5kg what is the percentage yield?
% yield = actual yield x 100
theoretical yield
= 65.3kg
x 100
70.5kg
= 92.6%