Oleum also known as fuming sulphuric acid has the molecular formula (\(H_2S_2O_7\)). It can be considered as a mixture od \(SO_3\) and \(H_2SO_4\). Addition of water to Oleum converts the free \(SO_3\) into \(H_2SO_4\) and the resulting solution will contain only \(H_2SO_4\).
\[SO_3 + H_2O \rightarrow H_2SO_4\]
To know how much water to add to a known amount of oleum, the concentration of oleum is expressed in terms of a percentage. The percentage will look like (100 + x)% and will always be more than 100. The value of “x“ is the amount of \(H_2O\) required by 100g of oleum to convert all the free \(SO_3\) present in it into \(H_2SO_4\).
For example, if a sample of oleum is labelled 109%, it means that to 100g of it 9g of water should be added to convert all the free \(SO_3\) present in it into \(H_2SO_4\).
\[SO_3 + H_2O \rightarrow H_2SO_4\]
According to the above reaction, 18g of \(H_2O\) is needed to convert 80g \(SO_3\) into \(H_2SO_4\). So, 9g of \(H_2O\) is needed to convert 40g \(SO_3\) into \(H_2SO_4\). This means that in 100g of the sample of oleum, 40g is \(SO_3\) and the rest 60g is \(H_2SO_4\).
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