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Chapter 22: Problem 9

Write a chemical equation for the hydrolysis of \(\mathrm{XeF}_{4}\) that yields \(\mathrm{XeO}_{3}, \mathrm{Xe}, \mathrm{O}_{2},\) and \(\mathrm{HF}\) as products.

Short Answer

Expert verified
The balanced chemical equation for the hydrolysis of \(XeF_4\) that yields \(XeO_3\), \(Xe\), \(O_2\), and \(HF\) is \n\n \(2XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + 6HF\)

Step by step solution

01

Write Down the Unbalanced Equation

The unbalanced chemical equation for the hydrolysis of \(XeF_4\) should be written as follows:\n\n\(XeF_4 + H_2O \rightarrow XeO_3 + Xe + O_2 + HF\)
02

Balance the Oxygen Atoms

Begin balancing the equation starting from the Oxygen atoms on both sides. As there are three Oxygen atoms in \(XeO_3\) and two in \(O_2\), add three water molecules to balance the reactant side. The equation thus becomes:\n\n\(XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + HF\)
03

Balance the Hydrogen and Fluorine Atoms

Six Hydrogen atoms in the reactant side must be balanced by adding six molecules of \(HF\) at the product side. This also balances out the Fluorine atoms. The equation then becomes:\n\n\(XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + 6HF\)
04

Balance the Xenon Atoms

Finally balance the Xenon atoms. As there are two Xenon atoms on the product side (where one is from \(XeO_3\) and the other is from \(Xe\)), add one more molecule of \(XeF_4\) to make two Xenon atoms on the reactant side. This leads to the final balanced equation:\n\n\(2XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + 6HF\)

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Most popular questions from this chapter

In \(1988,\) G. J. Schrobilgen, professor of chemistry at McMaster University in Canada, reported the synthesis of an ionic compound, \([\mathrm{HCNKrF}]\left[\mathrm{AsF}_{6}\right],\) which consists of \(\mathrm{HCNKr} \mathrm{F}^{+}\) and \(\mathrm{AsF}_{6}^{-}\) ions. In the \(\mathrm{HCNKr} \mathrm{F}^{+}\) ion, the krypton is covalently bonded to both fluorine and nitrogen. Draw Lewis structures for these ions, and estimate the bond angles.

In the extraction of bromine from seawater (reaction 22.3), seawater is first brought to a pH of 3.5 and then treated with \(\mathrm{Cl}_{2}(\mathrm{g}) .\) In practice, the \(\mathrm{pH}\) of the seawater is adjusted with \(\mathrm{H}_{2} \mathrm{SO}_{4},\) and the mass of chlorine used is \(15 \%\) in excess of the theoretical. Assuming a seawater sample with an initial pH of 7.0 a density of \(1.03 \mathrm{g} \mathrm{cm}^{-3}\), and a bromine content of 70 ppm by mass, what masses of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{Cl}_{2}\) would be used in the extraction of bromine from \(1.00 \times 10^{3} \mathrm{L}\) of seawater?

The structures of the \(\mathrm{NH}_{3}\) and \(\mathrm{NF}_{3}\) molecules are similar, yet the dipole moment for the \(\mathrm{NH}_{3}\) molecule is rather large (1.47 debye) and that of the NF \(_{3}\) molecule is rather small (0.24 debye). Provide an explanation for this difference in the dipole moments.

To displace \(\mathrm{Br}_{2}\) from an aqueous solution of \(\mathrm{Br}^{-}\) add \((\mathrm{a}) \mathrm{I}_{2}(\mathrm{aq}) ;\) (b) \(\mathrm{Cl}_{2}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2}(\mathrm{g}) ;\) (d) \(\mathrm{Cl}^{-}(\mathrm{aq})\) (e) \(\mathrm{I}_{3}^{-}(\mathrm{aq})\).

Explain why the volumes of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) obtained in the electrolysis of water are not the same.
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