Question

Question #ef835

Answer 1

It is stated in the problem `UF_6` reacts with water to form a gas containing 95% F and 5% H along with a solid compound of U,F and O.
This compound is formed due to replacement of F atom by O atom.
The oxidation state of O is -2 and that of F is -1 in the compound.So each O atom will replace 2 F atoms.If x atom of O is present in the compound formed then F atom in the compound will be `(6-2x)`.
Let the molecuar formula of the formed by the action of water on `UF_6` is `UF_(6-2x)O_x`

Considering atomic masses as

`U->235"g/mol"`

`F->19"g/mol"`

`O->16"g/mol"`

`H->1"g/mol"`

The percentage composition of constituent elements in the gas `H-5% and F-95%` So the ratio of number of atoms of H and F in the compound
`H:F=5/1:95/19=5:5=1:1`

So the compound is HF (Hydrogen fluoride)

So the balanced equation of the reaction may be written as

`UF_6+xH_2O->UF_(6-2x)O_2 +2xHF(g)`

It is given that 3.730g `UF_(6-2x)O_x` is formed from 4.267g `UF_6`

So `("molar mass of "UF_(6-2x)O_x)/("molar mass of "UF_6)=3.730/4.267`

`=>(235+19(6-2x)+16x)/(235+19xx6)=3730/4267`

`=>349-22x=349xx(3730/4267)~~305`

`=>22x=349-305=44`

`=>x=2`

Hence MF of the compound formed

`UF_((6-2*2))O_2=color(red)(UF_2O_2)`

The blanced equation becomes

`UF_6+2H_2O->UF_2O_2 +4HF(g)`

It is evident from above equation that among 6 F atoms in original compound `UF_6` 2 atoms go to form the solid molecule of `UF_2O_2` and 4 atoms go to form four `HF(g)` molecules.

Hence `2/6=1/3=33.3%` goes to form the solid compound and `4/6=2/3=66.7%` goes to form gaseous molecules of HF.