Stoichiometry is the study of the relationships which exist between reactants and products in a chemical reaction. This relationship can easily be seen in a balanced chemical equation. For example,
K2SO4
------ 2K+ + SO42-
1 2 1
In the ionization reaction above, notice that
one-mole potassium tetraoxosulphate (VI) gives two moles of potassium ions and
one mole of tetraoxosulphate (VI) ion in solution. This ionization equation
also shows the mole ratio of the reactants and the products, which is known as the stoichiometry of the chemical reaction. It is necessary to note that the mole of a
substance is not affected if the reactants or products are ions, atoms or
molecules etc. in a chemical reaction.
CaCO3(s) + 2HCl(aq) ------ CaCl2(s) + H2O(l) + CO(g)
1mole
2mole 1mole 1 mole
1mole
Consider the chemical reaction above, and note the
relationship between the reactants and the products in moles. The reacting
masses of the above reaction can be calculated, since, the molar mass of a
substance is the mass of one mole of the substance expressed in grams.
Mole = reacting mass/molar mass
Then reacting mass = mole x molar mass
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Example
question 1
Calculate the mass of lead chloride that can be
obtained from 30g of lead trioxonitrate (V) in the presence of excess
hydrochloric acid. (Pb=207, Cl =35.5,N= 14, O=16, H=1)
Pb(NO3)2(s)
+ 2HCl(aq) -------- PbCl2(s) + HNO3(aq)
Solution
Pb(NO3)2(s) + 2HCl(aq)
-------- PbCl2(s) + HNO3(aq)
1mole 1mole
331g/mol 278g/mol
From
the equation
The reacting mass of lead trioxonitrate (V) =
1mole x 331g/mol = 331g
The mass of lead chloride that would produce = 1mole
x 278g/mol = 278g
331g of Pb(NO3)2 = 278g of PbCl2
30g of Pb(NO3)2 = 278÷331 x30
= 25.2g of PbCl2
Or
By
considering the number of moles involved in the reaction
1mole of
Pb(NO3)2 = 1mole of PbCl2
Then find the number of moles in 30g of Pb(NO3)2
Mole = reacting mass/molar mass
Reacting mass given = 30g
Molar mass = 331g/mol
Mole = 30/331 = 0.09063mol
Since 1mole of Pb(NO3)2 =
1mole of PbCl2
then
0.09063mole of Pb(NO3)2 = 0.09063mole of PbCl2
Mass in 0.09063mole of PbCl2 = 0.09063 x 278 =
25.2g
Example
question 2
What mass of silver trioxonitrate (V) would be
required to yield 14g of silver chloride in addition of excess sodium chloride?
(Na=23, Cl = 35.5,Ag=108, N=14,O=16)
NaCl(aq)
+ AgNO3(aq) --------- NaNO3(aq) + AgCl(s)
Solution
NaCl(aq) + AgNO3(aq)
--------- NaNO3(aq) + AgCl(s)
1mole 1mole
170g/mol 143.5g/mol
From
the equation
Reacting mass of AgNO3 = 1mole x
170g/mol = 170g
Mass of AgCl produced = 1mole x 143.5g/mol =
143.5g
143.5g of AgCl required = 170g of AgNO3
14g of AgCl will require = 170/143.5 x 14/1 = 16.6g
Or
By
considering the number of moles involved in the reaction
NaCl(aq) + AgNO3(aq)
--------- NaNO3(aq) + AgCl(s)
1mole
1mole
From
the equation
1mole of AgCl
= 1 mole of AgNO3
Number of moles in 14g of AgCl =
Reacting mass given = 14g
Molar mass
= 143.5g/mol
Mole = 14/143.5
=0.0975mole
Since
1mole of AgCl = 1mole of AgNO3
then
0.0975mole of AgCl = 0.0975mole of AgNO3
So, mass of AgNO3 in
0.0975 = 0.0975 x 170 =16.6g
Example
question 3
Calculate the mass of magnesium chloride that will
be formed, when 20g of magnesium is added to excess dilute hydrochloric acid. (
Mg = 24, Cl=35.5, H=1)
Mg(s)
+ 2HCl(aq) ----------- MgCl2(aq) + H2(g)
Solution
Mg(s)
+ 2HCl(aq) ----------- MgCl2(aq) + H2(g)
1mole
1mole
24g/mol
95g/mol
From
the equation
Reacting mass of Mg = 24g/mol x 1mole =24g
Mass of MgCl2 produced = 95g/mol x
1mole= 95g
If 24g of Mg = 95g of MgCl2
Then
20g of Mg = 95/24 x 20/1 = 79.2 g of MgCl2
Or
By
considering the number of moles involved in the reaction
1mole of Mg
= 1mole of MgCl2
Reacting mass given = 20g
Molar mass of Mg = 24g/mol
Number of moles in 20g of Mg = 20/24 = 0.8333mole
Since
1mole of Mg = 1mole of MgCl2
Then
0.8333mole of MgCl2 =
0.8333 mole of Mg
Mass of MgCl2 in 0.8333mole =
0.8333 x 95 =79.2g of MgCl2
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