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Lecture 4:
Stoichiometry
Contents
Preamble
Basics of stoichiometry
Ideal gas law
Excess and limiting reactants
Oxidation and reduction reactions
Conclusion
References
Key words: stoichiometry, ideal gas law, Dalton’s law, mole fraction
Preamble
Many of the materials balance problems that arise in metal extraction processes can be solved easily with either.
a) Knowledge of atomic/molecular weights of reactants and products appearing in a balanced chemical reaction, or b) From mass of compound contained in a given volume of gas.
Stoichiometry can be used to develop the above relationship.
Basics of stoichiometry
i) Atomic or molecular mass of an element and compound. Molecular weight of a compound is the addition of molecular weight (Atomic weight) of elements comprising the compound. For example, ൌ molecular 65.34 32 97.34 or ൎ 97 (^) ୩ ୫୭୪ୣ୩
Molecular weight of ZnS weight of Zn molecular weight of S ൌ ൌ or (^) ୫୭୪ୣ^ .
Similarly molecular weight of Feଶ O^ ଷ ൌ 160^ ୩ ୫୭୪ୣ୩
ii) Thechemical equation: law of conservation of mass
For any chemical reaction, the number of atoms of elements in the product = No. of atoms
To avoid confusion and error particularly in thermodynamic calculations, physical forms of reactants and
(s) solid, (l) liquid and (g) gas
2 Feሺsሻ (^) ଶ
of that elements in the reactants.
products should also be mentioned in parentheses of chemical symbols e.g.
ଷ ሺO
ଶ ሻ^ ൌ Feଶ O^ ଷ ሺsሻ
Cuଶ Sሺlሻ O (^) ଶ ሺgሻ ൌ 2Cuሺlሻ SO (^) ଶ ሺgሻ
Balanced equation shows
ଶ ଶ of^ SO^ ଶ
ଶto^2 moles^ of^ Cu^ and^1 mole^ of^ SO^ ଶ.
40% FeS is converted to liquid Cu and iron oxide
400Kg Fe S ൌ 4.551 Kg mole
The balanced chemical equatio is
S O (^) ଶ
Moles of SO (^) ଶ produced ൌ 8.32 Kg moles
Ideal gas law:
Most gases involved in metallurgical processes at high temperature follow ideal gas behavior.
PV ൌ nRT (1)
In the equation 1 P is pressure, T is temperature and n is moles of gas. 1 Kg mole gas at ሺ1.01325 ൈ 10 ହ^ atm. p 22.4 mଷ^ and 1 lb mole of gas ൌ 359ft ଷ^.
୬
- 1 mole of Cu S combines with 1 moles of O to give 2 moles of Cu and 1 mole
- 2 moles of Cu atoms in Cuଶ S from 2 moles of liquid Cu.
- 1 mole of S in Cuଶ S forms 1 mole of SO (^) ଶ
- 1 mole of O (^) ଶ forms 1 mole of SO (^) ଶ
- 1 mole of Cu (^) ଶ S requires 1 moleof O form
How may moles of SO (^) ଶ are produced when 1 ton of Cu matte containing 60%Cuଶ S and
600Kg Cuଶ S ൌ 3.77Kg moles
s
n
3.77 Cuଶ S 3.77 O (^) ଶ ൌ 7.54Cu 3.
resssure 273Kሻ ൌ
V (^) ൌ RT P (2)
At any given T and P volume/ mole or no of mole in a given volume of gas is same.
In gas mixtures made up of ideal gases, each component of the mixture obeys the ideal gas law. For i ୲୦^ component of an ideal mixture contained in a volume V at T
XCO మ ൌ VCOమV౪ ൌ 0.33; VHమOV౪ ൌ X H మ O ൌ 0.
hus, in ideal gases partial volume fraction of a component is same.
cess and limiting reactan
The exact amount of the reactants can be determined by the stoichiometry of the reaction. In many cases one or more reactants is supplied in cess to complete the reaction. Reactant present in the smallest stoichiometric amount is called the limiting reactant. Reactants supplied in excess of the
ୡୣୱୱ ୫୭୪ୣୱ ୰ୣ୯ୢ.୭୰ ୡ୭୫୮୭୬ୣ୬୲ୱ ୰ୣୟୡ୲୧୭୬
T pressure and
Ex ts
ex
limiting reactant are called excess reactants.
% excess ൌ ୫୭୪ୣୱ ୧୬ ୣ ୶^ ൈ 100
Reduction: Gain of electrons by an atom
Oxidation and reduction always occurs simultaneously so that electrons are conserved.
eous solutionሻ ՜ Cuሺsሻ
ሺୟ୯ ୱ୭୪.ሻ ሺୟ୯.ୱ୭୪ሻ
In this lecture basics of stoichiometry are discussed. The concepts are illustrated with suitable examples. The reader may more problems to develop skills in stoichiometric calculations. Some problems are discussed in lect re 5.
References
huhmann, JR. Metallurgical engineering, volume 1
Oxidation‐ Reduction Reactions.
Oxidation: Loss of electrons from an atom
Cementation reaction Cuଶା^ ሺin aqu
Cuଶା^ 2eത ൌ Cu
Fe ൌ Feଶା^ 2eത
Cuଶା^ ሺFeୱ ሻ ൌ Cuሺୱሻ Feଶା^.
Conclusion
solve u
- R. Sc
- Geiger: energy balance in metallurgical processes
