Question

Complete and balance each of the following double-replacement reactions: (a) \(\mathrm{MgSO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow\) (b) \(\operatorname{AlBr}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow\)

Short answer

(a) \(\mathrm{MgSO}_{4} + \mathrm{BaCl}_{2} \rightarrow \mathrm{MgCl}_{2} + \mathrm{BaSO}_{4}\) (b) \(2 \operatorname{AlBr}_{3} + 3 \mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow \mathrm{Al}_{2}(\text{CO}_{3})_{3} + 6 \operatorname{NaBr}\)

Step-by-step solution

Identify the Products for Reaction (a)

In double-replacement reactions, the cations and anions of two different compounds exchange places. Here, MgSO₄ and BaCl₂ are the reactants. Thus, the potential products will be MgCl₂ and BaSO₄.

Write the Balanced Equation for Reaction (a)

Now, knowing the products, write the equation: \[\mathrm{MgSO}_{4}(a q) + \mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{MgCl}_{2}(a q) + \mathrm{BaSO}_{4}(s) \]Notice that BaSO₄ is a solid precipitate, while MgCl₂ remains in aqueous form. Check the formula and the balancing: Each side has 1 Mg, 1 Ba, 2 Cl, and 1 SO₄ group, so the equation is balanced.

Identify the Products for Reaction (b)

In the second reaction, AlBr₃ and Na₂CO₃ are the compounds involved. The products will be Al₂(CO₃)₃ and NaBr, following the double-displacement rule.

Write the Unbalanced Equation for Reaction (b)

The reaction can be written as: \[\operatorname{AlBr}_{3}(a q) + \mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Al}_{2}( ext{CO}_{3})_{3}(s) + \operatorname{NaBr}(a q) \]. Notice Al₂(CO₃)₃ is a solid as it precipitates out of solution.

Balance the Equation for Reaction (b)

To balance, we need to ensure the same number of each type of atom on both sides of the equation. This means adjusting coefficients: \[2 \operatorname{AlBr}_{3}(a q) + 3 \mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Al}_{2}(\text{CO}_{3})_{3}(s) + 6 \operatorname{NaBr}(a q) \]. Now, both sides have 2 Al, 6 Br, 6 Na, 3 CO₃ groups balance, ensuring the equation is balanced.

Key concepts

Chemical Equation Balancing

The process of chemical equation balancing ensures that the number of each type of atom is the same on both sides of a chemical equation. In a chemical reaction, atoms are neither created nor destroyed; they are rearranged. Balancing equations maintains this principle of the conservation of mass, which is crucial for accurately representing what happens in the reaction.

For double-replacement reactions like the ones given, this means ensuring that the sum of the atoms of reactants equals the sum of the atoms in the products. For instance, when balancing Reaction (a) with \( \text{MgSO}_{4}(aq) + \text{BaCl}_{2}(aq) \rightarrow \text{MgCl}_{2}(aq) + \text{BaSO}_{4}(s) \), checking the atom types reveals 1 magnesium, 1 barium, 2 chlorine, and 1 sulfate group on each side.

If any of these counts don't match on the reactant and product sides, you would adjust the coefficients (the numbers in front of molecules), ensuring that the total number of each type of atom is identical. This can involve some trial and error initially, but practice makes the process more intuitive. It is good practice to double-check your balanced equation to confirm that all components are correctly accounted for.

Precipitation Reactions

Precipitation reactions occur when two solutions react and form an insoluble product, called a precipitate. These reactions are visually identifiable by the formation of a solid from what was initially solutions that appeared clear.

In the given reactions, look at Reaction (a): \(\text{MgSO}_{4}(aq) + \text{BaCl}_{2}(aq) \rightarrow \text{MgCl}_{2}(aq) + \text{BaSO}_{4}(s) \). Here, barium sulfate (\(\text{BaSO}_{4}(s)\)) emerges as a solid precipitate out of the aqueous solutions. This signifies a precipitation reaction.

We find precipitates by consulting solubility rules, which specify groups of ions that usually form soluble or insoluble compounds. For example, while most sulfate compounds are soluble, barium sulfate is an exception, which is why it precipitates. Watching for these patterns helps predict and understand classroom demonstration outcomes, or when conducting experiments yourself.

Aqueous Solutions

Aqueous solutions are key to reactions in liquid mediums, where the solutes are dissolved in water. Understanding this concept helps identify the state of matter for reactants and products during reactions. The symbol \((aq)\) denotes substances that are dissolved in water, meaning the ions are free to move and react.

In our given reactions, substances like \(\text{MgSO}_{4}(aq)\) and \(\text{BaCl}_{2}(aq)\) are in the aqueous state, allowing them to interact through their dissolved ions. For example, the ions in \(\text{MgSO}_{4}(aq)\) separate, moving freely in solution and can immediately react with ions from \(\text{BaCl}_{2}(aq)\) to produce an insoluble compound like \(\text{BaSO}_{4}(s)\).

The characterization of substances in such reactions is critical, as studying aqueous solutions sheds light on various phenomena ranging from biological processes to industrial reactions. This understanding makes it possible to predict outcomes and balance equations in double-replacement, precipitation, and other reactions efficiently.