How are xenon fluorides `XeF_(2)`, `XeF_(4)` and `XeF_(6)` obtained?

Explanation

`XeF_(2)`, `XeF_(4)`, and `XeF_(6)` are obtained by a direct reaction between `Xe` and `F_(2)`. The condition under which the reaction is carried out determines the product.
`underset(("Excess"))(Xe_((g))) + F_(2(g))overset("673K, 1bar")to XeF_(2(s))`
`underset(("1 : 5 ratio"))(Xe_((g))) + 2F_(2(g))overset("873K, 7bar")to XeF_(4(s))`
`underset(("1 : 20 ratio"))(Xe_((g))) + 3F_(2(g))overset("573K,60-70bar")to XeF_(6(s))`