Question #ef835

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Question #ef835

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Answer 1 · 2016-10-03T08:12:54

It is stated in the problem $U F_{6}$ $UF_6$ 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 $(6 - 2 x)$ $(6-2x)$ .
Let the molecuar formula of the formed by the action of water on $U F_{6}$ $UF_6$ is $U F_{6 - 2 x} O_{x}$ $UF_(6-2x)O_x$

Considering atomic masses as

$U \to 235 g/mol$ $U->235"g/mol"$

$F \to 19 g/mol$ $F->19"g/mol"$

$O \to 16 g/mol$ $O->16"g/mol"$

$H \to 1 g/mol$ $H->1"g/mol"$

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

So the compound is HF (Hydrogen fluoride)

So the balanced equation of the reaction may be written as

$U F_{6} + x H_{2} O \to U F_{6 - 2 x} O_{2} + 2 x H F (g)$ $UF_6+xH_2O->UF_(6-2x)O_2 +2xHF(g)$

It is given that 3.730g $U F_{6 - 2 x} O_{x}$ $UF_(6-2x)O_x$ is formed from 4.267g $U F_{6}$ $UF_6$

So $\frac{molar mass of U F_{6 - 2 x} O_{x}}{molar mass of U F_{6}} = \frac{3.730}{4.267}$ $("molar mass of "UF_(6-2x)O_x)/("molar mass of "UF_6)=3.730/4.267$

$\Rightarrow \frac{235 + 19 (6 - 2 x) + 16 x}{235 + 19 \times 6} = \frac{3730}{4267}$ $=>(235+19(6-2x)+16x)/(235+19xx6)=3730/4267$

$\Rightarrow 349 - 22 x = 349 \times (\frac{3730}{4267}) \approx 305$ $=>349-22x=349xx(3730/4267)~~305$

$\Rightarrow 22 x = 349 - 305 = 44$ $=>22x=349-305=44$

$\Rightarrow x = 2$ $=>x=2$

Hence MF of the compound formed

$U F_{(6 - 2 \cdot 2)} O_{2} = U F_{2} O_{2}$ $UF_((6-2*2))O_2=color(red)(UF_2O_2)$

The blanced equation becomes

$U F_{6} + 2 H_{2} O \to U F_{2} O_{2} + 4 H F (g)$ $UF_6+2H_2O->UF_2O_2 +4HF(g)$

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

Hence $\frac{2}{6} = \frac{1}{3} = 33.3 %$ $2/6=1/3=33.3%$ goes to form the solid compound and $\frac{4}{6} = \frac{2}{3} = 66.7 %$ $4/6=2/3=66.7%$ goes to form gaseous molecules of HF.

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