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Chapter 22: Problem 14

Fluorine can be prepared by the reaction of hexafluoromanganate(IV) ion, MnF \(_{6}^{2-}\), with antimony pentafluoride to produce manganese(IV) fluoride and \(\mathrm{SbF}_{6}^{-}\) followed by the disproportionation of manganese(IV) fluoride to manganese(III) fluoride and \(\mathrm{F}_{2}(\mathrm{g}) .\) Write chemical equations for these two reactions.

Short Answer

Expert verified
Balanced equations: \[ \mathrm{MnF_{6}^{2-}} + 2\mathrm{SbF_{5}} \rightarrow \mathrm{MnF_{4}} + 2\mathrm{SbF_{6}^{-}} \] and \[ 2\mathrm{MnF_{4}} \rightarrow 2\mathrm{MnF_{3}} + \mathrm{F_{2}(\mathrm{g})}\]

Step by step solution

01

Write Reactants and Products for Reaction 1

Reaction 1 involves hexafluoromanganate(IV) ion, or \( \mathrm{MnF_{6}^{2-}} \), and antimony pentafluoride, or \( \mathrm{SbF_{5}} \), reacting to form manganese(IV) fluoride, or \( \mathrm{MnF_{4}} \), and \(\mathrm{SbF_{6}^{-}}\). Write them as follows: \( \mathrm{MnF_{6}^{2-}} + \mathrm{SbF_{5}} \rightarrow \mathrm{MnF_{4}} + \mathrm{SbF_{6}^{-}} \)
02

Balance Reaction 1

On the right hand side, there are six fluorine atoms in \(\mathrm{SbF_{6}^{-}}\) and four in \(\mathrm{MnF_{4}}\), a total of ten. On the left hand side, there are five fluorine atoms in \(\mathrm{SbF_{5}}\) and six in \(\mathrm{MnF_{6}^{2-}}\), a total of eleven. Thus, the reaction is unbalanced. To balance, we add a coefficient of 2 to \(\mathrm{SbF_{5}}\). The balanced equation becomes: \( \mathrm{MnF_{6}^{2-}} + 2\mathrm{SbF_{5}} \rightarrow \mathrm{MnF_{4}} + 2\mathrm{SbF_{6}^{-}} \)
03

Write Reactants and Products for Reaction 2

Reaction 2 involves the disproportionation of manganese(IV) fluoride, or \( \mathrm{MnF_{4}} \), to produce manganese(III) fluoride, or \( \mathrm{MnF_{3}} \), and \(\mathrm{F_{2}(\mathrm{g})}\). Write them as follows: \( \mathrm{MnF_{4}} \rightarrow \mathrm{MnF_{3}} + \mathrm{F_{2}(\mathrm{g})}\)
04

Balance Reaction 2

In this reaction, there are four fluorine atoms on the left but only three plus two on the right. Add a coefficient of 2 in front of \( \mathrm{MnF_{4}} \) and \( \mathrm{MnF_{3}}\) on the left and right respectively. The balanced equation becomes: \( 2\mathrm{MnF_{4}} \rightarrow 2\mathrm{MnF_{3}} + \mathrm{F_{2}(\mathrm{g})}\)

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Most popular questions from this chapter

On the basis of molecular orbital theory, would you expect \(\mathrm{NH}_{2}^{+}\) to be linear or bent?

Zn can reduce \(\mathrm{NO}_{3}^{-}\) to \(\mathrm{NH}_{3}(\mathrm{g})\) in basic solution. (The following equation is not balanced.) $$\begin{aligned}\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})+& \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \\\&\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{g})\end{aligned}$$ The \(\mathrm{NH}_{3}\) can be neutralized with an excess of \(\mathrm{HCl}(\mathrm{aq})\) Then, the unreacted HCl can be titrated with NaOH. In this way a quantitative determination of \(\mathrm{NO}_{3}^{-}\) can be achieved. A 25.00 mL sample of nitrate solution was treated with zinc in basic solution. The \(\mathrm{NH}_{3}(\mathrm{g})\) was passed into \(50.00 \mathrm{mL}\) of \(0.1500 \mathrm{M} \mathrm{HCl} .\) The excess \(\mathrm{HCl}\) required \(32.10 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for its titration. What was the \(\left[\mathrm{NO}_{3}\right]\) in the original sample?

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Which of the following reactions are likely to go to completion or very nearly so? (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1)\) (b) \(\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1)+4 \mathrm{Cl}^{-}(\mathrm{aq}) \stackrel{-}{\longrightarrow}\) \(2 \mathrm{Cl}_{2}(\mathrm{g})+4 \mathrm{OH}^{-}(\mathrm{aq})\) (c) \(\mathrm{O}_{3}(\mathrm{g})+\mathrm{Pb}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(\mathrm{PbO}_{2}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})\) (d) \(\mathrm{HO}_{2}^{-}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(3 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(1)\)
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