Q115 E
Question
A mixture of Xenon and Flourine was heated. A sample of white solid that formed reacted with hydrogen to yield 81 mL of Xenon (at STP) and hydrogen fluoride , which was collected in water, giving a solution of hydrofluoric acid . The hydrofluoric acid solution was titrated , and 68.43 mL of 0.3172 M of Sodium Hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon .
Step-by-Step Solution
VerifiedThe white solid that formed must be a Xenon Flouride. The compound would be \({\rm{Xe}}{{\rm{F}}_{\rm{6}}}\).
Let compound be \({\rm{Xe}}{{\rm{F}}_{\rm{n}}}\)
We can write the empirical equation , \({\rm{2Xe + n}}{{\rm{F}}_{\rm{2}}} \to {\rm{2Xe}}{{\rm{F}}_{\rm{n}}}\) ; \({\rm{2Xe}}{{\rm{F}}_{\rm{n}}}{\rm{ + n}}{{\rm{H}}_{\rm{2}}} \to {\rm{2Xe + 2nHF}}\)
Mole of Xe formed = \(\frac{{{\rm{81}}}}{{{\rm{22400}}}}{\rm{ = 3}}{\rm{.61 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\)
[ At STP ,the volume of 1mole of gas = 22400 ml ]
Mole of HF formed = \(\frac{{{\rm{68}}{\rm{.43 \times 0}}{\rm{.3172}}}}{{{\rm{1000}}}}{\rm{ = 0}}{\rm{.0217}}\)
∴ \(\frac{{{\rm{MoleofXe}}}}{{{\rm{MoleofHF}}}} = \frac{{\rm{2}}}{{{\rm{2n}}}}\)
Or, \(\)\(\frac{{{\rm{3}}{\rm{.61 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}}}{{{\rm{0}}{\rm{.0217}}}}{\rm{ = }}\frac{{\rm{2}}}{{{\rm{2n}}}}\)
Or, n = 6
So, the empirical formula for the white solid is \({\rm{Xe}}{{\rm{F}}_{\rm{6}}}\) and the balanced chemical equations are
\({\rm{2Xe + 6}}{{\rm{F}}_{\rm{2}}} \to {\rm{2Xe}}{{\rm{F}}_{\rm{6}}}\); \({\rm{2Xe}}{{\rm{F}}_{\rm{6}}}{\rm{ + 6}}{{\rm{H}}_{\rm{2}}} \to {\rm{2Xe + 12HF}}\)