Problem 59 - Chemistry and Chemical Reactivity

Question

Balance the following equations, and then classify each as a precipitation, acid-base, or gas-forming reaction. (a) \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2 (\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{CoCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\) (c) \(\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+\mathrm{NaNO}_{3}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) Acid-base, (b) Gas-forming, (c) Precipitation.

Step 1: Balance Equation for Reaction (a)

The initial unbalanced equation is \( \mathrm{Ba(OH)}_{2}(\mathrm{aq}) + \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) \). To balance:1. Balance Ba: 1 Ba atom is already present on both sides.2. Balance Cl: Add 2 moles of HCl to get 2 Cl atoms on both sides.3. Balance H: 2 OH and 2 HCl gives a total of 4 H atoms, which balances with 2 molecules of H2O.4. The balanced equation is \( \mathrm{Ba(OH)}_{2}(\mathrm{aq}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq}) + 2\mathrm{H}_{2}\mathrm{O}(\ell) \). This is an acid-base reaction because it involves the reaction of a base \( \mathrm{Ba(OH)}_{2} \) with an acid \( \mathrm{HCl} \).

Step 2: Balance Equation for Reaction (b)

The initial unbalanced equation is \( \mathrm{HNO}_{3}(\mathrm{aq}) + \mathrm{CoCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g}) \). To balance:1. Balance Co: One Co atom is present on both sides.2. Balance NO3: Add 2 moles of HNO3 to get 2 NO3 groups on the product side.3. Balance H: 2 H from 2 HNO3 balances with 1 H2O.4. Balance C: 1 CO3 goes to form 1 CO2.5. Balance O: The oxygen is balanced with the other atoms.6. The balanced equation is \( 2\mathrm{HNO}_{3}(\mathrm{aq}) + \mathrm{CoCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g}) \).This is a gas-forming reaction because it produces \( \mathrm{CO}_{2} \) gas.

Step 3: Balance Equation for Reaction (c)

The initial unbalanced equation is \( \mathrm{Na}_{3}\mathrm{PO}_{4}(\mathrm{aq}) + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) + \mathrm{NaNO}_{3}(\mathrm{aq}) \).To balance:1. Balance Cu: Place 3 before Cu(NO3)2 and 3 Cu on the product side.2. Balance PO4: Place 2 before Na3PO4 to have adequate PO4 ions on both sides.3. Balance Na and NO3: This leads to 6 NaNO3 to balance Na and NO3.4. The balanced equation is \( 2\mathrm{Na}_{3}\mathrm{PO}_{4}(\mathrm{aq}) + 3\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) + 6\mathrm{NaNO}_{3}(\mathrm{aq}) \).This is a precipitation reaction because \( \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2} \) precipitates out of the solution.

Key Concepts

Acid-Base Reaction

Chemical reactions can take many forms, and one common type is the acid-base reaction. In such reactions, an acid and a base interact to form a salt and water.
This process is also known as neutralization and is fundamental to many chemical processes in nature and industry.

• Acid: A substance that donates protons (H⁺) in a reaction. In our exercise, the acid is hydrochloric acid, HCl.
• Base: A substance that accepts protons. Here, our base is barium hydroxide, Ba(OH)₂, which donates OH^- ions.
• Products: The products of an acid-base reaction typically include a salt (BaCl₂) and water (H₂O).

Balancing acid-base reactions involves ensuring equal numbers of all types of atoms on both sides of the equation, paying close attention to the protons and the hydroxide ions involved.
For instance, in the given reaction, two moles of HCl are needed to react with one mole of Ba(OH)₂ to achieve balance. This forms two moles of water and one mole of barium chloride.

Gas-Forming Reaction

Gas-forming reactions are intriguing chemical processes where a gas is one of the main products. These reactions often involve carbonates or bicarbonates reacting with acids.
This yields carbon dioxide, a gas, among other products.

• Reactants: Typically include acid and a carbonate compound. In our reaction, nitric acid (HNO₃) reacts with cobalt carbonate (CoCO₃).
• Gas Formation: The reaction releases carbon dioxide gas (CO₂), a hallmark of gas-forming reactions.
• Other Products: Besides the gas, water (H₂O) and a salt (Co(NO₃)₂) are also formed.

Balancing gas-forming reactions requires accounting for all atoms, ensuring that the moles of gases released are correctly represented.
In the example, two moles of nitric acid are required per mole of cobalt carbonate to balance the equation and generate one mole each of water and CO₂.

Precipitation Reaction

Precipitation reactions occur when two aqueous solutions react to form an insoluble solid known as the precipitate.
This type of reaction is commonly seen in environmental and industrial applications to remove unwanted substances from solutions.

• Reactants: Typically involve two aqueous solutions of ionic compounds. Here, sodium phosphate (Na₃PO₄) reacts with copper(II) nitrate (Cu(NO₃)₂).
• Precipitate Formation: The products include copper(II) phosphate (Cu₃(PO₄)₂), which precipitates out of the solution.
• Soluble Products: Other products, such as sodium nitrate (NaNO₃), remain in solution.

Balancing precipitation reactions involves ensuring charge and mass balance, focusing on the ions that stay dissolved and those that form the precipitate.
In the balanced equation given, three moles of copper(II) nitrate and two moles of sodium phosphate react, producing one mole of copper(II) phosphate. This showcases the typical stoichiometry of such reactions.