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Chapter 4: Problem 26

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) \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{FeCl}_{3}(a q) \longrightarrow\) (b) $\mathrm{ZnCl}_{2}(a q)+\mathrm{Cs}_{2} \mathrm{CO}_{3}(a q) \longrightarrow$ (c) $\mathrm{Na}_{2} \mathrm{~S}(a q)+\operatorname{CoSO}_{4}(a q) \longrightarrow$

Short Answer

Expert verified
(a) The balanced net ionic equation for case (a) is: \(2OH^-(aq) + Fe^{3+}(aq) \longrightarrow 2Fe(OH)_3(s)\). The spectator ions are \(Ba^{2+}\) and \(Cl^-\). (b) The balanced net ionic equation for case (b) is: \(Zn^{2+}(aq) + CO_3^{2-}(aq) \longrightarrow ZnCO_3(s)\). The spectator ions are \(Cs^+\) and \(Cl^-\). (c) The balanced net ionic equation for case (c) is: \(S^{2-}(aq) + Co^{2+}(aq) \longrightarrow CoS(s)\). The spectator ions are \(Na^+\) and \(SO_4^{2-}\).

Step by step solution

01

Write Molecular Equation - a

Write the molecular equation for the given reaction: \(Ba(OH)_2(aq) + FeCl_3(aq) \longrightarrow\)
02

Predict Products and Balance the Equation - a

In this reaction, we will have a double replacement reaction. The products will be \(BaCl_2\) and \(Fe(OH)_3\). Balancing the equation, we get: \(Ba(OH)_2(aq) + FeCl_3(aq) \longrightarrow BaCl_2(aq) + 2Fe(OH)_3(s)\)
03

Dissociate Strong Electrolytes - a

Now dissociate all strong electrolytes in the reaction: \(Ba^{2+}(aq) + 2OH^-(aq) + Fe^{3+}(aq) + 3Cl^-(aq) \longrightarrow Ba^{2+}(aq) + 2Cl^-(aq) + 2Fe(OH)_3(s)\)
04

Cancel Spectator Ions - a

Identify and cancel spectator ions in this reaction. The spectator ions are \(Ba^{2+}\) and \(Cl^-\), so: \(2OH^-(aq) + Fe^{3+}(aq) \longrightarrow 2Fe(OH)_3(s)\) (b)
05

Write Molecular Equation - b

Write the molecular equation for the given reaction: \(ZnCl_2(aq) + Cs_2CO_3(aq) \longrightarrow\)
06

Predict Products and Balance the Equation - b

In this reaction, we will have a double replacement reaction. The products will be \(ZnCO_3\) and \(CsCl\). Balancing the equation, we get: \(ZnCl_2(aq) + Cs_2CO_3(aq) \longrightarrow ZnCO_3(s) + 2CsCl(aq)\)
07

Dissociate Strong Electrolytes - b

Now dissociate all the strong electrolytes in the reaction: \(Zn^{2+}(aq) + 2Cl^-(aq) + 2Cs^+(aq) + CO_3^{2-}(aq) \longrightarrow ZnCO_3(s) + 2Cs^+(aq) + 2Cl^-(aq)\)
08

Cancel Spectator Ions - b

Identify and cancel spectator ions in this reaction. The spectator ions are \(Cs^+\) and \(Cl^-\), so: \(Zn^{2+}(aq) + CO_3^{2-}(aq) \longrightarrow ZnCO_3(s)\) (c)
09

Write Molecular Equation - c

Write the molecular equation for the given reaction: \(Na_2S(aq) + CoSO_4(aq) \longrightarrow\)
10

Predict Products and Balance the Equation - c

In this reaction, we will have a double replacement reaction. The products will be \(Na_2SO_4\) and \(CoS\). Balancing the equation, we get: \(Na_2S(aq) + CoSO_4(aq) \longrightarrow Na_2SO_4(aq) + CoS(s)\)
11

Dissociate Strong Electrolytes - c

Now dissociate all the strong electrolytes in the reaction: \(2Na^+(aq) + S^{2-}(aq) + Co^{2+}(aq) + SO_4^{2-}(aq) \longrightarrow 2Na^+(aq) + SO_4^{2-}(aq) + CoS(s)\)
12

Cancel Spectator Ions - c

Identify and cancel spectator ions in this reaction. The spectator ions are \(Na^+\) and \(SO_4^{2-}\), so: \(S^{2-}(aq) + Co^{2+}(aq) \longrightarrow CoS(s)\)

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