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Chapter 21: Problem 74

Show that, in principle, \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) can be converted almost completely to NaOH(aq) by the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) $$

Short Answer

Expert verified
Yes, in theory and ideally, Na2CO3 can be converted almost entirely into NaOH using the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) + \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{CaCO}_{3}(\mathrm{s}) + 2 \mathrm{NaOH}(\mathrm{aq})\), assuming there's sufficient Ca(OH)2 available.

Step by step solution

01

Recognize the Reaction Type

This is a double replacement (also known as metathesis) chemical reaction where two compounds exchange ions or bonds to form different compounds. Specifically, Na2CO3 is reacting with Ca(OH)2 to form CaCO3 and NaOH.
02

Analyze the Stoichiometry

Looking at the balanced chemical equation, it can be seen that 1 mole of Na2CO3 reacts with 1 mole of Ca(OH)2 to produce 1 mole of CaCO3 and 2 moles of NaOH. Here, we must note that Na2CO3 is converted almost completely to NaOH, which means Na2CO3 is the limiting reagent in this reaction.
03

Understand the Reaction Conditions

Assuming we have enough Ca(OH)2, then every Na2CO3 molecule will react and will form two NaOH molecules for every one Na2CO3 molecule.
04

Result

Therefore, in principle, Na2CO3(aq) can be converted almost completely to NaOH(aq) by the reaction with Ca(OH)2(s). The Na2CO3 will all be used up, resulting in 2 times as many moles of NaOH, which is the final product of the reaction.

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Most popular questions from this chapter

Lithium superoxide, \(\mathrm{LiO}_{2}(\mathrm{s}),\) has never been isolated. Use ideas from Chapter \(12,\) together with data from this chapter and Appendix \(D\), to estimate \(\Delta H_{f}\) for \(\mathrm{LiO}_{2}(\mathrm{s})\) and assess whether \(\mathrm{LiO}_{2}(\mathrm{s})\) is thermodynamically stable with respect to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}).\) (a) Use the Kapustinskii equation, along with appropriate data below, to estimate the lattice energy, \(U,\) for \(\left.\mathrm{LiO}_{2}(\mathrm{s}) . \text { (See exercise } 126 \text { in Chapter } 12 .\right)\) The ionic radii for \(L\) i \(^{+}\) and \(O_{2}^{-}\) are \(73 \mathrm{pm}\) and \(144 \mathrm{pm},\) respectively. (b) Use your result from part (a) in the BornFajans-Haber cycle to estimate \(\Delta H_{\mathrm{f}}^{2}\) for \(\mathrm{LiO}_{2}(\mathrm{s})\) [Hint: For the process \(\mathrm{O}_{2}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}_{2}^{-}(\mathrm{g}), \Delta H^{\circ}=.\) \(-43 \mathrm{kJ} \mathrm{mol}^{-1} .\) See Table 21.2 and Appendix \(\mathrm{D}\) for the other data that are required.] (c) Use your result from part (b) to calculate the enthalpy change for the decomposition of \(\mathrm{LiO}_{2}(\mathrm{s})\) to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) For \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s}), \Delta H_{\mathrm{f}}^{\circ}=-598.73\) \(\mathrm{kJmol}^{-1}.\) (d) Use your result from part (c) to decide whether \(\mathrm{LiO}_{2}(\mathrm{s})\) is thermodynamically stable with respect to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) Assume that entropy effects can be neglected.

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