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Chapter 4: Problem 24

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow\).

Short Answer

Expert verified
(a) The net ionic equation is: \(2\mathrm{Cr}^{3+}(a q)+2\mathrm{CO}_{3}^{2-}(a q) \longrightarrow\mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s)\). The spectator ions are \(\mathrm{NH}_{4}^{+}(a q)\) and \(\mathrm{SO}_{4}^{2-}(a q)\). (b) The net ionic equation is: \(\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{BaSO}_{4}(s)\). The spectator ions are \(\mathrm{K}^{+}(a q)\) and \(\mathrm{NO}_{3}^{-}(a q)\). (c) The net ionic equation is: \(\mathrm{Fe}^{2+}(a q)+2\mathrm{OH}^{-}(a q) \longrightarrow\mathrm{Fe}(\mathrm{OH})_{2}(s)\). The spectator ions are \(\mathrm{K}^{+}(a q)\) and \(\mathrm{NO}_{3}^{-}(a q)\).

Step by step solution

01

(a) Predict the products of the reaction

For the given reaction: \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) We can predict the products by exchanging the cations and anions: Chromium will combine with carbonate ion, and ammonium will combine with sulfate ion. So the products will be chromium(III) carbonate (Cr2(CO3)3) and ammonium sulfate ((NH4)2SO4).
02

(a) Write the balanced molecular equation

The balanced molecular equation for this reaction is: $$ \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+2\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s)+2(\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}(a q) $$
03

(a) Write the total ionic equation

The total ionic equation is obtained by breaking all the aqueous compounds into their respective ions: $$ 2\mathrm{Cr}^{3+}(a q)+3\mathrm{SO}_{4}^{2-}(a q)+4\mathrm{NH}_{4}^{+}(a q)+2\mathrm{CO}_{3}^{2-}(a q) \longrightarrow\mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s)+4\mathrm{NH}_{4}^{+}(a q)+2\mathrm{SO}_{4}^{2-}(a q) $$
04

(a) Write the net ionic equation

The net ionic equation is obtained by canceling out the spectator ions found on both sides of the total ionic equation. In this case, the spectator ions are \(\mathrm{NH}_{4}^{+}(a q)\) and \(\mathrm{SO}_{4}^{2-}(a q)\). $$ 2\mathrm{Cr}^{3+}(a q)+2\mathrm{CO}_{3}^{2-}(a q) \longrightarrow\mathrm{Cr}_{2}(\mathrm{CO}_{3})_{3}(s) $$
05

(b) Predict the products of the reaction

For the given reaction: \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2}\mathrm{SO}_{4}(a q) \longrightarrow\) We can predict the products by exchanging the cations and anions: Barium will combine with sulfate ion, and potassium will combine with nitrate ion. So the products will be barium sulfate (BaSO4) and potassium nitrate (KNO3).
06

(b) Write the balanced molecular equation

The balanced molecular equation for this reaction is: $$ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2}\mathrm{SO}_{4}(a q) \longrightarrow \mathrm{BaSO}_{4}(s)+2\mathrm{KNO}_{3}(a q) $$
07

(b) Write the total ionic equation

The total ionic equation is obtained by breaking all the aqueous compounds into their respective ions: $$ \mathrm{Ba}^{2+}(a q)+2\mathrm{NO}_{3}^{-}(a q)+2\mathrm{K}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow\mathrm{BaSO}_{4}(s)+2\mathrm{K}^{+}(a q)+2\mathrm{NO}_{3}^{-}(a q) $$
08

(b) Write the net ionic equation

The net ionic equation is obtained by canceling out the spectator ions found on both sides of the total ionic equation. In this case, the spectator ions are \(\mathrm{K}^{+}(a q)\) and \(\mathrm{NO}_{3}^{-}(a q)\). $$ \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{BaSO}_{4}(s) $$
09

(c) Predict the products of the reaction

For the given reaction: \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow\) We can predict the products by exchanging the cations and anions: Iron will combine with hydroxide ion, and potassium will combine with nitrate ion. So the products will be iron(II) hydroxide (Fe(OH)2) and potassium nitrate (KNO3).
10

(c) Write the balanced molecular equation

The balanced molecular equation for this reaction is: $$ \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{KOH}(a q) \longrightarrow\mathrm{Fe}(\mathrm{OH})_{2}(s)+2\mathrm{KNO}_{3}(a q) $$
11

(c) Write the total ionic equation

The total ionic equation is obtained by breaking all the aqueous compounds into their respective ions: $$ \mathrm{Fe}^{2+}(a q)+2\mathrm{NO}_{3}^{-}(a q)+2\mathrm{K}^{+}(a q)+2\mathrm{OH}^{-}(a q) \longrightarrow\mathrm{Fe}(\mathrm{OH})_{2}(s)+2\mathrm{K}^{+}(a q)+2\mathrm{NO}_{3}^{-}(a q) $$
12

(c) Write the net ionic equation

The net ionic equation is obtained by canceling out the spectator ions found on both sides of the total ionic equation. In this case, the spectator ions are \(\mathrm{K}^{+}(a q)\) and \(\mathrm{NO}_{3}^{-}(a q)\). $$ \mathrm{Fe}^{2+}(a q)+2\mathrm{OH}^{-}(a q) \longrightarrow\mathrm{Fe}(\mathrm{OH})_{2}(s) $$

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Most popular questions from this chapter

(a) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of some salt and then spill some of it. What happens to the concentration of the solution left in the container? (b) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) aqueous solution of some salt and let it sit out, uncovered, for a long time, and some water evaporates. What happens to the concentration of the solution left in the container? (c) A certain volume of a \(0.50 \mathrm{M}\) solution contains \(4.5 \mathrm{~g}\) of a salt. What mass of the salt is present in the same volume of a \(2.50 \mathrm{M}\) solution?

Determine the oxidation number of sulfur in each of the following substances: (a) barium sulfate, \(\mathrm{BaSO}_{4},\) (b) sulfurous acid, \(\mathrm{H}_{2} \mathrm{SO}_{3},\) (c) strontium sulfide, \(\mathrm{SrS},\) (d) hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S} .\) (e) Based on these compounds what is the range of oxidation numbers seen for sulfur? Is there any relationship between the range of accessible oxidation states and sulfur's position on the periodic table?

Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+},\) and \(\mathrm{Fe}^{2+},\) which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+}\). (a) If \(1500 \mathrm{~L}\) of hard water contains \(0.020 \mathrm{M} \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+}\), how many moles of \(\mathrm{Na}^{+}\) are needed to replace these ions? (b) If the sodium is added to the water softener in the form of \(\mathrm{NaCl}\), how many grams of sodium chloride are needed?

A solution is made by mixing \(15.0 \mathrm{~g}\) of \(\mathrm{Sr}(\mathrm{OH})_{2}\) and \(55.0 \mathrm{~mL}\) of \(0.200 \mathrm{MHNO}_{3}\). (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resultant solution acidic or basic?

We have seen that ions in aqueous solution are stabilized by the attractions between the ions and the water molecules. Why then do some pairs of ions in solution form precipitates? \([\) Section 4.2\(]\)
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