If oxygen is 20.9 vol % of air then LFL for n-hexane [tex]C_{6} H_{14} .[/tex]is 1.1percent LFL is half the stochiometric amount so LFL is 1.1 percent.
When carbon reacts with hydrogen and oxygen then it will form carbon di-oxide and water.
[tex]2C_{6} +H_{14} +19O_{2}[/tex] ⟶[tex]12CO_{2} +14H_{2} O[/tex]
Concentration needed for complete combustion is calculated through the ratio of our compound and oxygen. To make it easier, let`s imagine we have 100 moles of air. That means we have 20.9 moles of [tex]O_{2}[/tex]
Stochiometric amount of hexane needed to react with that amount of [tex]O_{2}[/tex] would be 20.9 *2 /19=2.2 mol .
If oxygen is 20.9 vol % of air then LFL for n-hexane [tex]C_{6} H_{14} .[/tex]is 1.1percent
LFL is half the stochiometric amount so LFL is 1.1 percent.
To know more about the Stochiometric here
https://brainly.com/question/12912444
#SPJ4