# I cannot understand the reactions?

1. Apr 6, 2017

### Vengo

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???

2. Apr 6, 2017

### Comeback City

They have different proportions of each compound. In first reaction, PbS : O2 ratio is 2:3, but it is 1:2 in second reaction.

3. Apr 6, 2017

### Vengo

Then it is about stoichiomtery which determines the reaction ????

4. Apr 6, 2017

### Curiosity 1

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. :-)

5. Apr 6, 2017

### Drakkith

Staff Emeritus
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.

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.

6. Apr 7, 2017

### DrDu

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.

7. Apr 7, 2017

### Staff: Mentor

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted