On the formation of `SF_(6)` by the direct combination of S and `F_(2)` which is the limiting reagent? Prove it.

Explanation

`SF_(6)` is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of S is allowed to react with 12 moles of Fluorine.
`S_((l))+3F_(2_((g))) rarr SF_(6_((g)))`
As per the stoichiometric reaction, one mole of S reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.
`:.` It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.