Stoichiometry and chemical reaction
The study of the quantitative relationships implied by chemical reactions is called stoichiometry. A chemical equation is both a quantitative and a qualitative statement of a chemical change. A balanced chemical equation, expressed a number of laws and facts. These laws are those of; (a) conservation of mass and (b) constant ( definite) composition.
Note: stoichiometry is a term used by chemists and chemical engineers.
A stoichiometric diagram of the combustion reaction of methane
Mole ratios and mass
The numerical coefficients of a balanced equation represent the numbers of moles of reactants and products. From these coefficients, we get the mole ratio of the reactants and products in a reaction. e.g
H2SO4(aq) > 2H+(aq) + SO2-4(aq)
Of moles. 1. 2 1
Mole ratio. 1. : 2 : 1
Note that one mole of any element contains the same number of atoms. This number, which has a value of 6.02 × 1023 is called the Avogadro number or constant. Chemists often express the concentration of a solution in moles per cubic decimeter (mol dm-3). This concentration is called molar concentration (m). Number of moles = mass/molar mass
1.Calculation based on chemical equations
Considering the equation
2NO(g) + O2(g) > 2NO2(g)
The quantitative information provided is that, two moles of nitrogen (ii) oxide reacts with one mole of oxygen to produce two moles of nitrogen (IV) oxide. Using molar masses, 60g of nitrogen (ii) oxide, since the three substances are gases and at standard temperature and pressure, one mole of molecules of any gas occupies 22.4dm3. Therefore, 44.8dm3 of nitrogen (ii) oxide reacts with 22.4dm3 of oxygen to yield 44.8dm3 of nitrogen (IV) oxide.
2. Calculations involving gas volume
4NH3(g) + 5O2(g) > 4NO(g) + 6H2O(g) if 250cm3 of NH3 are burned completely, what volume of (a) oxygen is used up? (b) NO is produced?
Solution: equation of reaction
4NH3(g) + 5O2(g) > 4NO(g) + 6H2O(g)
Volume. 4 : 5 : 4 : 6
Ratio. 1 5/4 1 6/4
250cm3 of NH3 will used up 5/4 × 250
= 312.5cm3 of oxygen
250cm3 of NH3 will produce 1×250 = 250cm3 of NO.
3. Calculations involving liquid volume
What volume of 1.0M HCL will be required to react completely with 3.25g of zinc.
Solution: equation of reaction
Zn + 2HCL > ZnCL2 + H2
From the equation
65g of Zn react with 73g of HCL :. 3.25g of Zn will react with 73 × 3.25g/65 of HCL
= 3.65g of HCL
Converting this mass of HCL to a volume of 1.0M solution 3.65g is contained in 1000cm3 of 1.0M solution. :. 3.65g of HCL will be contained in 100 × 3.65cm3/3.65
= 100cm3 of 1.0m HCL
:. The volume of 1.0 HCL requires is 100cm3
4. Calculations involving masses
Calculate the mass of carbon (IV) oxide produced on burning 105g of ethyne (C=12, O=16, H=1)
2C2H2(g) + 5O2(g) > 4CO2(g) + 2H2O(g)
Mass: 26g 44g
Ratio: 2 : 4
Number of moles of C2H2 reacts = 105/26
From the equation, 2moles of C2H2 produces 4 moles of CO2 :. 4moles of C2H2 will produce 4×4 mole of CO2
2 = 8moles
Reacting mass = No of moles X molar mass = 8×44=352g of CO2
Thus, 352g of CO2 is produced by burning 104g of CH.
Volumetric Analysis usually includes titrations of acid against base or trioxocarbonate (IV), oxidizing agent against reducing agent or one substance against another giving a precipitate.
Indicators for acid-base titrations
|Acids and base||Indicator|
|Strong acid and strong base||Any indicator|
|Strong acid and weak base or trioxocarbonate IV||Methyl orange|
|Weak acid and strong base||Phenolphthalein|
|Weak acid and weak base||No suitable indicator|
Definition of terms used in volumetric Analysis
1.Mass concentration: the mass concentration of a solution is the amount of solute present in a given volume of the solution. It is expressed in g/dm3 or gdm-3 thus, mass concentration= molar concentration × molar mass
2. Molar concentration: it is defined as the concentration of a solution in moles per dm3.
Number of mole of a substance
= Number of particles
6.02 × 1023
Number of moles of a substance
= Mass of substance mol-1/ Molar mass
Note: in volume per cm3 of solution, we have. Number of moles of solute
= Volume × molar concentration/ 1000
I.e = V/1000 × M mole
3. Standard solution: a standard solution is a solution of a known concentration.
4. Molar solution: a molar solution of a compound is one which contains one mole or the molar mass of the compound in one dm3 of the solution.
1. How many miles of (H+) are there in 1dm3 of 0.5 M solution of H2SO4? (Jamb 1992)
Solution: H2SO4 > 2H+ + SO2-4
1mole of H2SO4 contains 2 mole of H+ :. (0.5×1000/1000) mole of H2SO4 will contain (2/1 × 0.5) of H+ = 1.0 mole
2. A balance chemical equation obeys the law of? (Jamb 1993)
A. Conversation of Mass
B. Definite proportions
C. Multiple proportions
D. Conservation of energy
3. What volume of carbon(ii)oxide is produced by reacting excess carbon with 10dm3 of oxygen? (Jamb 2013)
Solution: 2C(s) + O2(g) > 2CO(g)
1vol O2 produced 2 vol CO
10dm3 will produce 2/1 × 10dm3=5dm3
4.what is the percentage of sulphur in sulphur (IV) oxide? (Jamb 2014)
D. 50% [S=32, O=16]
Solution: SO2 = 32 + (2×16)
=32 + 32 =64; 32/64 × 100% = 50%
5. Define molar solution?
Answer: a molar solution of a compound is one which contains one mole or the molar mass of the compound in one dm3 of the solution.
6. Mention the two laws that are required in a balanced equation?
Answer: conservation of mass and constant composition
7. Name the two careers/jobs that make use of stoichiometry?
Answer: chemists and chemical engineers
8. Defined stoichiometry?
Answer: it is defined as the study of quantitative relationships implied by chemical reactions.