I cannot understand the reactions?

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SUMMARY

Chemical reactions are determined by stoichiometry and the specific conditions under which they occur. For example, the reactions 2PbS + 3O2 → 2 PbO + 2SO2 and PbS + 2O2 → PbSO4 illustrate how different ratios of reactants lead to different products. The stoichiometric ratios dictate the formation of products, with the first reaction requiring a 2:3 ratio of PbS to O2 for PbO production, while the second requires a 1:2 ratio for PbSO4. Additionally, the actual yield of reactions can differ from the theoretical yield due to side reactions, emphasizing the complexity of reaction conditions.

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Vengo
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Actually, I want to how the chemical reactions are written. Is it based on stoichiometry or atomic configuration or just by assumption, For eg.
2PbS + 3O2 → 2 PbO + 2SO2
PbS + 2O2 → PbSO4
look at these reactions they have same compounds but different products. How is it happening?
 
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They have different proportions of each compound. In first reaction, PbS : O2 ratio is 2:3, but it is 1:2 in second reaction.
 
Then it is about stoichiomtery which determines the reaction ?
 
Yes, if you want to get PbO, then you need to mix PbS and O2 in the ratio of 2:3. If you want to get PbSO4, then you need to mix PbS and O2 in the ratio of 1:2. :-)
 
Multiple reaction products may be produced from the same reactants depending on the exact conditions and the nature of the reactants.
Reactions are often written as if they are "ideal". In other words, they are written as if those other reactions cannot occur. But this is just a simplification.

Vengo said:
Actually, I want to how the chemical reactions are written. Is it based on stoichiometry or atomic configuration

Under the maximum "theoretical yield", they are one and the same. But the actual yield is often less than the theoretical yield due to side reactions.
 
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The question under which conditions which products will form is a very complicated one. It is better understood in the case of reactions involving organic compounds than for inorganic ones. A well known example is the reaction of toluene (methylbenzene) with Cl2. If a catalyst is present, Cl2 will be split into Cl##^+## and Cl##^-## and the Cl##^+## will attack the benzene ring. Without catalyst and at higher temperatures or when energy is provided by an intensive light source, the Cl2 will be split into two Cl atoms which attack preferentially the methyl group.
 
Curiosity 1 said:
Yes, if you want to get PbO, then you need to mix PbS and O2 in the ratio of 2:3. If you want to get PbSO4, then you need to mix PbS and O2 in the ratio of 1:2. :-)

No, that's not how it works. Even if you mix PbS and O2 in the exact 2:3 ratio nothing can stop them from reacting according to the second reaction equation just leaving excess unreacted PbS.
 
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