Excess `F_(2)`(g) reacts at `150^(@)C` and 1.0 atm pressure with `Br_(2)`(g) to give a compound `BrF_(n)` .If 423 mL of `Br_(2)`(g) at the same temperature and pressure produced 4.2 g of `BrF_(n)` what is n? (Br=80, F= 19)

Options

A. 3
B. 1
C. 5
D. 7

Answer

C

Explanation

`w(Br_2) = (pVM)/(RT) = (1.0 xx 0.423 xx 160)/(0.082 xx 423) = 1.95`
`because 1.95` g of `Br_(2)` produced 4.2 g `BrF_(n)` , 80 g `Br_2` will produce `(4.2)/(1.95) xx 80 = 172.2 g Br F_(n)`
`implies 80 + 19 n = 172.2 or n = 4.85 implies` Formula = `BrF_(5)`