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Chapter 22: Problem 50

What mass of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) was present in a sample that required \(26.50 \mathrm{mL}\) of \(0.0510 \mathrm{M} \mathrm{KMnO}_{4}\) for its oxidation to \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in an acidic solution? \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\).

Short Answer

Expert verified
The mass of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) in the sample is approximately 0.42 g.

Step by step solution

01

Balancing the Redox Reaction

The above redox reaction can be balanced in an acidic medium as follows: \[5 \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+2 \mathrm{KMnO}_{4}(aq)+14 \mathrm{H}^{+}(aq) \rightarrow 5 \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+2 \mathrm{Mn}^{2+}(aq)+8 \mathrm{H}_{2} \mathrm{O}(l)\] Here, 5 moles of sodium sulfite react with 2 moles of potassium permanganate.
02

Find Moles of KMnO4

The moles of KMnO4 can be calculated using the formula: \[Moles = Molarity \times Volume\] So, moles of KMnO4 = \(0.0510 \mathrm{M} \times 26.50 \times 10^{-3} \mathrm{L} = 0.001345 \mathrm{mol}\)
03

Calculate Moles of Na2SO3

From the stoichiometry of the reaction, we know that 5 moles of Na2SO3 react with 2 moles of KMnO4. So, moles of Na2SO3 = \( \frac{5}{2}\) moles of KMnO4 = \( \frac{5}{2} \times 0.001345 \mathrm{mol} = 0.0033625 \mathrm{mol}\)
04

Calculate Mass of Na2SO3

Finally, we calculate the mass of Na2SO3 using its molar mass (126.04 g/mol): Mass = Moles × Molar Mass = \(0.0033625 \mathrm{mol} \times 126.04 \mathrm{g/mol} = 0.42 g\)

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Most popular questions from this chapter

Each of the following compounds produces \(\mathrm{O}_{2}(\mathrm{g})\) when strongly heated: (a) \(\mathrm{HgO}(\mathrm{s}) ;\) (b) \(\mathrm{KClO}_{4}(\mathrm{s})\) (c) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{s}) ;\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) .\) Write a plausible equation for the reaction that occurs in each instance.

The sketch is a portion of the phase diagram for the element sulfur. The transition between solid orthorhombic \(\left(S_{\alpha}\right)\) and solid monoclinic \(\left(S_{\beta}\right)\) sulfur, in the presence of sulfur vapor, is at \(95.3^{\circ} \mathrm{C} .\) The triple point involving monoclinic sulfur, liquid sulfur, and sulfur vapor is at \(119^{\circ} \mathrm{C}\) (a) How would you modify the phase diagram to represent the melting of orthorhombic sulfur that is sometimes observed at \(113^{\circ} \mathrm{C} ?[\) Hint: What would the phase diagram look like if the monoclinic sulfur did not form? (b) Account for the observation that if a sample of rhombic sulfur is melted at \(113^{\circ} \mathrm{C}\) and then heated, the liquid sulfur freezes at \(119^{\circ} \mathrm{C}\) upon cooling.

Use Lewis structures and other information to explain the observation that (a) the oxygen-to-oxygen bond lengths in \(\mathrm{O}_{2}, \mathrm{O}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\) are \(121,128,\) and \(148 \mathrm{pm},\) respectively. (b) the oxygen-to-oxygen bond length of \(\mathrm{O}_{2}\) is \(121 \mathrm{pm}\) and for \(\mathrm{O}_{2}^{+}\) is \(112 \mathrm{pm}\). Why is the bond length for \(\mathrm{O}_{2}^{+}\) so much shorter than for \(\mathrm{O}_{2} ?\)

Predict the geometric structures of (a) \(\mathrm{BrF}_{3} ;\) (b) IF \(_{5}\); (c) \(\mathrm{Cl}_{3} \mathrm{IF}^{-}\). (Central atom underlined.).

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.
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