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Balance each equation and write the corresponding net ionic equation. $$ \begin{aligned} &\text { (a) } \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{NaOH}(a q) \rightarrow \\ &\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{NaNO}_{3}(a q) \\ &\text { (b) } \mathrm{Zn}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{ZnS}(s)+\mathrm{KClO}_{4}(a q) \\ &\text { (c) } \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \rightarrow \\ &\mathrm{Pb}(\mathrm{OH})_{2}(s)+\mathrm{KNO}_{s}(a q) \\ &\text { (d) } \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow \\ &\mathrm{ZnCO}_{3}(s)+\mathrm{NaNO}_{s}(a q) \end{aligned} $$

Short Answer

Expert verified
(a) Fe(OH)_3: Fe^(3+) + 3OH^-. (b) ZnS: Zn^(2+) + S^(2-). (c) Pb(OH)_2: Pb^(2+) + 2OH^-. (d) ZnCO_3: Zn^(2+) + CO_3^(2-).

Step by step solution

01

Balance Equation (a)

For the reaction \(\text{Fe(NO}_3\text{)}_3 + \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaNO}_3\), we need to balance the number of each type of atom on both sides. - There are 1 Fe on each side.- On the left, there are 3 NO鈧冣伝 ions and 3 Na for the NaOH, so we need 3 NaNO鈧 on the right.- This gives us 3 NaOH units on the left to match the 3 OH鈦 ions in \(\text{Fe(OH)}_3\) on the right.Final balanced equation:\[\text{Fe(NO}_3\text{)}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaNO}_3\]
02

Write Net Ionic Equation (a)

In the balanced reaction, identify and remove spectator ions (Na鈦 and NO鈧冣伝). The net ionic equation shows only the ions participating in the formation of the precipitate Fe(OH)鈧:\[\text{Fe}^{3+} (aq) + 3\text{OH}^- (aq) \rightarrow \text{Fe(OH)}_3 (s)\]
03

Balance Equation (b)

For \(\text{Zn(ClO}_4\text{)}_2 + \text{K}_2 \text{S} \rightarrow \text{ZnS} + 2\text{KClO}_4\), ensure each type of atom is balanced. - There is 1 Zn and 1 S, balanced as ZnS.- ClO鈧勨伝 has 2 on the left, so we need 2 KClO鈧 on the right.Balanced equation:\[\text{Zn(ClO}_4\text{)}_2 + \text{K}_2 \text{S} \rightarrow \text{ZnS} + 2\text{KClO}_4\]
04

Write Net Ionic Equation (b)

Remove spectator ions (K鈦 and ClO鈧勨伝) from the balanced equation. The net ionic equation for the formation of ZnS is:\[\text{Zn}^{2+} (aq) + \text{S}^{2-} (aq) \rightarrow \text{ZnS} (s)\]
05

Balance Equation (c)

For \(\text{Pb(NO}_3\text{)}_2 + 2\text{KOH} \rightarrow \text{Pb(OH)}_2 + 2\text{KNO}_3\), balance the atoms:- 1 Pb on each side.- 2 NO鈧冣伝 needs 2 KNO鈧.- 2 K from KOH needs to be balanced on the right.Balanced equation:\[\text{Pb(NO}_3\text{)}_2 + 2\text{KOH} \rightarrow \text{Pb(OH)}_2 + 2\text{KNO}_3\]
06

Write Net Ionic Equation (c)

Identify spectator ions (K鈦 and NO鈧冣伝) and write the net ionic equation for the precipitate Pb(OH)鈧:\[\text{Pb}^{2+} (aq) + 2\text{OH}^- (aq) \rightarrow \text{Pb(OH)}_2 (s)\]
07

Balance Equation (d)

For \(\text{Zn(NO}_3\text{)}_2 + \text{Na}_2 \text{CO}_3 \rightarrow \text{ZnCO}_3 + 2\text{NaNO}_3\), follow these steps:- 1 Zn and 1 CO鈧 required, as ZnCO鈧 is formed.- Two NO鈧冣伝 in one Zn(NO鈧)鈧, so 2 NaNO鈧 needed.Balanced equation:\[\text{Zn(NO}_3\text{)}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{ZnCO}_3 + 2\text{NaNO}_3\]
08

Write Net Ionic Equation (d)

Spectator ions are Na鈦 and NO鈧冣伝. Remove these to give the net ionic equation for the formation of ZnCO鈧:\[\text{Zn}^{2+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{ZnCO}_3 (s)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Net Ionic Equation
The net ionic equation is a powerful concept in chemistry that simplifies reactions by eliminating spectator ions. These equations focus solely on the species that undergo a change during the reaction. This makes it easier to understand the core change happening in a chemical system.

To write a net ionic equation, start by writing the balanced chemical equation. Then, dissociate all strong electrolytes into their ions. Look at all ions present on each side of the equation and cancel out the spectator ions, which are ions that do not participate in the actual chemical change. What is left is the net ionic equation.

For example, in reaction (a), the balanced chemical equation is:\[\text{Fe(NO}_3\text{)}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaNO}_3\]Here, the spectator ions are \( \text{Na}^+ \) and \( \text{NO}_3^- \). Cancel out these ions to get the net ionic equation:\[\text{Fe}^{3+} (aq) + 3\text{OH}^- (aq) \rightarrow \text{Fe(OH)}_3 (s)\]

Using net ionic equations can help you focus on the essentials of chemical reactions, simplifying analysis and study.
Precipitation Reactions
Precipitation reactions are a fascinating type of chemical reaction where two soluble salts react in aqueous solutions to form an insoluble solid called the precipitate. These reactions are essential in various applications, such as in analytical chemistry, treatment of waste water, and even in everyday life such as in the formation of scale.

For a precipitation reaction to occur, the resulting product of the reaction must form an insoluble compound. In our examples, these are often metal hydroxides, sulfides, or carbonates that form precipitates given their low solubility in water. In reaction (a), \( \text{Fe(OH)}_3 \), a reddish-brown solid, precipitates out of the solution, resulting from the combination of \( \text{Fe}^{3+} \) ions and \( \text{OH}^- \) ions. This is typical of many metal hydroxides, which are insoluble.

To predict whether a precipitate will form, one can use solubility rules, which are guidelines that help determine the solubility of compounds in water. These rules help in predicting the outcome of mixing different ionic compounds in a solution.
Spectator Ions
Spectator ions are ions that appear on both the reactant and product side of a chemical equation and do not participate actively in the chemical reaction. They are "spectators" because they do not affect the outcome of the reaction but are present in the solution before and after the reaction occurs.

In our exercise, spectator ions include \( \text{Na}^+ \) and \( \text{NO}_3^- \) in reaction (a). These ions stay dissolved in the solution and remain unchanged in nature. Identifying and removing these ions when writing net ionic equations simplifies the equation, focusing only on the species involved in the actual chemical change.

Spectator ions serve as a good reminder of why comprehensive understanding of ionic equations is necessary. Ignoring them helps students concentrate on the reacting ions that contribute to the formation of the precipitate. Keeping track of these ions can also be useful in stoichiometric calculations where concentrations of ions are critical to the calculation of reaction yields or titrations.

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Most popular questions from this chapter

Vinegar, which is an aqueous solution of acetic acid, is used to remove deposits of calcium carbonate (lime scale) from automatic coffee makers. Write the equation describing the reaction between acetic acid and calcium carbonate.

Use the solubility rules to predict the products of the following reactions. In each case complete and balance the equation and write the net ionic equation. If no precipitate forms, then write "no reaction." (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}\left(\mathrm{ClO}_{4}\right)_{2}(a q) \rightarrow\) (b) \(\mathrm{AgNO}_{3}(a q)+\mathrm{NaClO}_{4}(a q) \rightarrow\) (c) \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}(a q) \rightarrow\) (d) \(\mathrm{Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+\mathrm{KBr}(a q) \rightarrow\)

Classify each of the reactions described by the following chemical equations as combination, decomposition, single replacement, or double replacement. (a) \(\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (b) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) (c) \(2 \mathrm{AgBr}(s)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{AgCl}(s)+\mathrm{Br}_{2}(l)\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+2 \mathrm{NaI}(a q) \rightarrow 2 \mathrm{AgI}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\)

Nitrogen forms oxyacids with formulas of \(\mathrm{HNO}_{2}\) and \(\mathrm{HNO}_{3}\), whereas phosphorous forms oxyacids with formulas of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and arsenic forms oxyacids with formulas of \(\mathrm{H}_{3} \mathrm{AsO}_{3}\) and \(\mathrm{H}_{3} \mathrm{AsO}_{4}\). Explain why nitrogen, which is in the same group as phosphorous and arsenic, does not form oxyacids analogous to those of the other two elements.

Write the formulas of the following common household chemicals: (a) sodium hypochlorite (the active ingredient bleach) (b) hydrogen peroxide (used medicinally in dilute solutions) (c) potassium hydroxide (found in some drain cleaners) (d) acetic acid (the active ingredient in vinegar)

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