Problem 52
Question
Balance the following equations, and then classify each as a precipitation, acid-base, or gas-forming reaction. (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})+\mathrm{KNO}_{3}(\mathrm{aq})\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{aq})\) (c) \(\mathrm{MgCO}_{3}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\)
Step-by-Step Solution
Verified Answer
(a) Precipitation; (b) Precipitation; (c) Gas-forming.
1Step 1: Balance Equation (a)
The unbalanced equation is \( \mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})+\mathrm{KNO}_{3}(\mathrm{aq}) \). To balance it, ensure that the number of each type of atom is equal on both sides:- Copper (Cu): 1 Cu on both sides- Carbon (C): 1 C on both sides- Oxygen (O): 3 O in CuCO₃, 3 O in one NO₃−, 3 O in KNO₃, and 3 O in KNO₃; balanced overall- Potassium (K): 2 K from K₂CO₃ on the left, 2 K needed on the right, thus 2 KNO₃- Nitrate (NO₃): 2 nitrate ions with Cu(NO₃)₂, thus 2 KNO₃ on the rightBalanced Equation: \( \mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})+2\mathrm{KNO}_{3}(\mathrm{aq}) \).
2Step 2: Classify Reaction (a)
In the balanced equation for (a), \( \mathrm{CuCO}_{3} \) is a solid precipitate forming from aqueous reactants. Therefore, this is a precipitation reaction.
3Step 3: Balance Equation (b)
The unbalanced equation is \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{aq}) \). Ensure quantities are equal on both sides:- Lead (Pb): 1 Pb on both sides- Chlorine (Cl): 2 Cl in PbCl₂ on the right, so 2 HCl on the left- Hydrogen (H): 2 H in 2 HCl- Nitrate (NO₃): 2 NO₃ in \( \mathrm{Pb}(\mathrm{NO}_{3})_{2} \) on the left, 2 needed on the right, thus 2 HNO₃Balanced Equation: \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s})+2\mathrm{HNO}_{3}(\mathrm{aq}) \).
4Step 4: Classify Reaction (b)
In the balanced equation for (b), \( \mathrm{PbCl}_{2} \) forms as a solid precipitate from aqueous reactants, indicative of a precipitation reaction.
5Step 5: Balance Equation (c)
The unbalanced equation is \( \mathrm{MgCO}_{3}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}\mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) \). Balance the equation:- Magnesium (Mg): 1 Mg on both sides- Carbon (C): 1 C on both sides- Chlorine (Cl): 2 Cl in MgCl₂ on the right, so 2 HCl needed on the left- Hydrogen (H): 2 H in 2 HCl- Oxygen (O): 3 O in CO₂ and H₂O, matching 3 O in MgCO₃Balanced Equation: \( \mathrm{MgCO}_{3}(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}\mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) \).
6Step 6: Classify Reaction (c)
In the balanced equation for (c), \( \mathrm{CO}_{2} \) gas is formed, indicating this is a gas-forming reaction.
Key Concepts
Precipitation ReactionsGas-Forming ReactionsAcid-Base Reactions
Precipitation Reactions
Precipitation reactions are a type of chemical reaction where two solutions combine to form an insoluble solid, known as a precipitate. Imagine mixing two clear liquids and suddenly seeing a solid form – that's precipitation at work! This happens when the ions in the solution come together to form a compound that does not dissolve in water.
In the examples given, like the reaction between potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)) and copper(II) nitrate (\(\mathrm{Cu(NO_3)_2}\)), a solid copper carbonate (\(\mathrm{CuCO}_3\)) forms as a precipitate. This reaction can be represented by the balanced equation:
In the examples given, like the reaction between potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)) and copper(II) nitrate (\(\mathrm{Cu(NO_3)_2}\)), a solid copper carbonate (\(\mathrm{CuCO}_3\)) forms as a precipitate. This reaction can be represented by the balanced equation:
- \( \mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq}) + \mathrm{Cu} \left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s}) + 2 \mathrm{KNO}_{3}(\mathrm{aq}) \)
Gas-Forming Reactions
Gas-forming reactions are fascinating because they result in the formation of a gas, often visible as bubbles. These reactions are particularly engaging as they demonstrate a clear transformation from solid or liquid reactants to a gaseous product.
In the exercise, magnesium carbonate (\(\mathrm{MgCO_3}\)) reacts with hydrochloric acid to yield magnesium chloride, water, and carbon dioxide gas. The balanced equation is:
In the exercise, magnesium carbonate (\(\mathrm{MgCO_3}\)) reacts with hydrochloric acid to yield magnesium chloride, water, and carbon dioxide gas. The balanced equation is:
- \( \mathrm{MgCO}_{3}(\mathrm{s}) + 2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g}) \)
Acid-Base Reactions
Acid-base reactions are fundamental in chemistry, involving the transfer of a proton (H⁺) from the acid to the base. Understanding these reactions is key to mastering many chemical processes. Although the original exercise didn't include a direct acid-base reaction example, let's explore how they generally work.
When an acid, such as hydrochloric acid (\(\mathrm{HCl}\)), reacts with a base, such as sodium hydroxide (\(\mathrm{NaOH}\)), they form water and salt. This neutralization can be represented by:
When an acid, such as hydrochloric acid (\(\mathrm{HCl}\)), reacts with a base, such as sodium hydroxide (\(\mathrm{NaOH}\)), they form water and salt. This neutralization can be represented by:
- \( \mathrm{HCl}(\mathrm{aq}) + \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{H}_{2}\mathrm{O}(\ell) + \mathrm{NaCl}(\mathrm{aq}) \)
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