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Chapter 5: Problem 107

When aqueous sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), is treated with dilute hydrochloric acid, HCl, the products are sodium chloride, water, and carbon dioxide gas. What is the net ionic equation for this reaction?

Short Answer

Expert verified
The net ionic equation for the reaction is: \(CO_3^{2-} + 2H^+ ➔ H_2O + CO_2\)

Step by step solution

01

Write the balanced molecular equation

First, start by writing down the balanced molecular equation for the reaction. Sodium Carbonate (\(Na_2CO_3\)) reacts with Hydrochloric acid (2\(HCl\)) to form Sodium Chloride (2\(NaCl\)), Water (\(H_2O\)) and Carbon dioxide (\(CO_2\)). So, the balanced molecular equation is: \(Na_2CO_3 + 2HCl ➔ 2NaCl + H_2O + CO_2\)
02

Break into ions

Next, dissociate all strong electrolytes into their ions. Sodium Carbonate (\(Na_2CO_3\)) breaks down into 2 Sodium ions (\(2Na^+\)) and one Carbonate ion (\(CO_3^{2-}\)). Hydrochloric acid (2\(HCl\)) breaks down into 2 Hydrogen ions (\(2H^+\)) and 2 Chloride ions (\(2Cl^-\)). Sodium Chloride (2\(NaCl\)) breaks down into 2 Sodium ions (\(2Na^+\)) and 2 Chloride ions (\(2Cl^-\)). Water and Carbon dioxide exist as molecules: \(H_2O\) and \(CO_2\). So, the complete ionic equation is:\(2Na^+ + CO_3^{2-} + 2H^+ + 2Cl^- ➔ 2Na^+ + 2Cl^- + H_2O + CO_2\)
03

Cross out spectator ions

Spectator ions appear on both sides of the equation and do not directly participate in the reaction. In this case, Sodium ions (\(Na^+\)) and Chloride ions (\(Cl^-\)) appear in their same form on both sides of the equation, so these are spectators and can be crossed out. What remains are the ions that reacted directly, forming the net ionic equation: \(CO_3^{2-} + 2H^+ ➔ H_2O + CO_2\)

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Most popular questions from this chapter

Explain why these reactions cannot occur as written. (a) \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow\) \(\mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\)

What are the oxidizing and reducing agents in the following redox reactions? (a) \(5 \mathrm{SO}_{3}^{2-}+2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+} \longrightarrow\) \(5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\) (b) \(2 \mathrm{NO}_{2}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+} \longrightarrow\) \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+2 \mathrm{H}_{2} \mathrm{O}\)

Balance these equations for redox reactions occurring in acidic solution. (a) $\mathrm{MnO}_{4}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{I}_{2}(\mathrm{s})$ (b) $\mathrm{BrO}_{3}^{-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{Br}^{-}+\mathrm{N}_{2}$ (c) $\mathrm{VO}_{4}^{3-}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{VO}^{2+}+\mathrm{Fe}^{3+}$ (d) $\mathrm{UO}^{2+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{UO}_{2}^{2+}+\mathrm{NO}(\mathrm{g})$

\(\mathrm{NH}_{3}(\mathrm{aq})\) conducts electric current only weakly. The same is true for \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) .\) When these solutions are mixed, however, the resulting solution is a good conductor. How do you explain this?

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).
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