If excess `F_(2)(g)` reacts at `0^(@)C` and 1.0 atm pressure with `Br_(2)(g)` to give a compound `BrF_(n)`, if 224ml of `Br_(2)(g)` at the same temperature and pressure produced 3.5g of `BrF_(n)`, what is n?

Explanation

`Br_(2)+nF_(2)rarr2BrF_(n)`
`224 ml Br_(2)` at 273 K and 1 atm
`n_(Br_(2))=((1)(224)xx10^(-3))/(0.0821xx273)=0.01` mol
`n_(Br)F_(n)=2n_(Br_(2))=0.02` mol
`(3.5)/((80+19n))=0.2`
3.5=1.6+0.38n
1.9=0.38n
`n=(1.9)/(0.38)=5`