Question ef835

Oct 3, 2016

It is stated in the problem $U {F}_{6}$ 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 $\left(6 - 2 x\right)$.
Let the molecuar formula of the formed by the action of water on $U {F}_{6}$ is $U {F}_{6 - 2 x} {O}_{x}$

Considering atomic masses as

$U \to 235 \text{g/mol}$

$F \to 19 \text{g/mol}$

$O \to 16 \text{g/mol}$

$H \to 1 \text{g/mol}$

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
$H : F = \frac{5}{1} : \frac{95}{19} = 5 : 5 = 1 : 1$

So the compound is HF (Hydrogen fluoride)

So the balanced equation of the reaction may be written as

$U {F}_{6} + x {H}_{2} O \to U {F}_{6 - 2 x} {O}_{2} + 2 x H F \left(g\right)$

It is given that 3.730g $U {F}_{6 - 2 x} {O}_{x}$ is formed from 4.267g $U {F}_{6}$

So $\left(\text{molar mass of "UF_(6-2x)O_x)/("molar mass of } U {F}_{6}\right) = \frac{3.730}{4.267}$

$\implies \frac{235 + 19 \left(6 - 2 x\right) + 16 x}{235 + 19 \times 6} = \frac{3730}{4267}$

$\implies 349 - 22 x = 349 \times \left(\frac{3730}{4267}\right) \approx 305$

$\implies 22 x = 349 - 305 = 44$

$\implies x = 2$

Hence MF of the compound formed

$U {F}_{\left(6 - 2 \cdot 2\right)} {O}_{2} = \textcolor{red}{U {F}_{2} {O}_{2}}$

The blanced equation becomes

$U {F}_{6} + 2 {H}_{2} O \to U {F}_{2} {O}_{2} + 4 H F \left(g\right)$

It is evident from above equation that among 6 F atoms in original compound $U {F}_{6}$ 2 atoms go to form the solid molecule of $U {F}_{2} {O}_{2}$ and 4 atoms go to form four $H F \left(g\right)$ 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.