Question

Complete and balance the following equations: (a) \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \stackrel{\text { spark }}{\longrightarrow}\) (b) \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) \stackrel{\Delta}{\longrightarrow}\) (c) \(\mathrm{HF}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\)

Short answer

(a) \( \text{C}_3\text{H}_6 + 4\text{O}_2 \rightarrow 3\text{CO}_2 + 3\text{H}_2\text{O} \), (b) \( \text{Na}_2\text{Cr}_2\text{O}_7 \cdot 2\text{H}_2\text{O} \rightarrow \text{Na}_2\text{O} + \text{Cr}_2\text{O}_3 + 2\text{H}_2\text{O} \), (c) \( 2\text{HF} + \text{Ca(OH)}_2 \rightarrow \text{CaF}_2 + 2\text{H}_2\text{O} \).

Step-by-step solution

Understand the type of reactions

Identify the types of reactions for each equation. Part (a) is a combustion reaction, part (b) is a decomposition reaction, and part (c) is a neutralization reaction.

Balance equation (a)

For the combustion of propene, the products are carbon dioxide (CO₂) and water (H₂O). The unbalanced equation is: \( \text{C}_3\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \). Balance carbons first: \( \text{C}_3\text{H}_6 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O} \). Then hydrogens: \( \text{C}_3\text{H}_6 + \text{O}_2 \rightarrow 3\text{CO}_2 + 3\text{H}_2\text{O} \). Finally, balance oxygens: \( \text{C}_3\text{H}_6 + 4\text{O}_2 \rightarrow 3\text{CO}_2 + 3\text{H}_2\text{O} \).

Decompose and balance equation (b)

\( \text{Na}_2\text{Cr}_2\text{O}_7 \cdot 2\text{H}_2\text{O} \) when heated decomposes into sodium oxide (Na₂O), chromium(III) oxide (Cr₂O₃), and water (H₂O). The unbalanced equation: \( \text{Na}_2\text{Cr}_2\text{O}_7 \cdot 2\text{H}_2\text{O}(s) \rightarrow \text{Na}_2\text{O} + \text{Cr}_2\text{O}_3 + 2\text{H}_2\text{O} \).

Balance equation (c)

The reaction of HF with \( \text{Ca(OH)}_2 \) forms calcium fluoride (CaF₂) and water. The unbalanced equation is: \( \text{HF} + \text{Ca(OH)}_2 \rightarrow \text{CaF}_2 + \text{H}_2\text{O} \). Balance hydrogens and fluorines: \( 2\text{HF} + \text{Ca(OH)}_2 \rightarrow \text{CaF}_2 + 2\text{H}_2\text{O} \).

Key concepts

Combustion Reaction

A combustion reaction involves the burning of a substance in the presence of oxygen, releasing energy in the form of heat or light. These reactions are usually identified by the reactant being a hydrocarbon, such as propene, combining with oxygen to produce carbon dioxide and water. In part (a) of this exercise, we see the combustion of propene, \( ext{C}_3 ext{H}_6 + ext{O}_2 \). Combustion reactions are exothermic, meaning they release energy. They are critical in everyday applications like running engines and generating electricity. To balance such reactions:

• Start by balancing the carbon atoms in the equation.
• Next, balance the hydrogen atoms.
• Finally, balance the oxygen atoms. You may need to adjust these as you go along because oxygen is present in both products.

In our equation, we first balanced the three carbon atoms from propene to form 3 carbon dioxide molecules. Then, we balanced the six hydrogen atoms to produce 3 water molecules. Finally, we adjusted the oxygen molecules to ensure there were enough to react completely with the hydrocarbons. As you practice, start looking for patterns like recognizing that in a complete combustion reaction, the products will always be carbon dioxide and water.

Decomposition Reaction

In decomposition reactions, a single compound breaks down into two or more simpler substances. Part (b) of our exercise reflects a decomposition reaction where \( ext{Na}_2 ext{Cr}_2 ext{O}_7 2 ext{H}_2 ext{O}(s) \) is heated to break it down into sodium oxide, chromium(III) oxide, and water. These reactions are typically endothermic, which requires the input of energy, often in the form of heat, to break the bonds. When balancing decomposition reactions:

• Identify the products that are formed, typically by looking for known stable compounds like oxides and carbonates.
• Write the unbalanced equation based on your understanding of the reactants.
• Adjust coefficients to ensure that all atoms on both sides of the equation are balanced.

For our specific problem, we write the unbalanced decomposition of sodium dichromate with water, then balance it to reflect the conservation of matter. Because each molecule of water contributes to the final balance of hydrogen and oxygen in the products, careful attention to detail is necessary.

Neutralization Reaction

Neutralization reactions are a type of chemical reaction where an acid and a base react to form water and a salt. This is demonstrated in part (c) of our exercise, where the reaction \( ext{HF} + ext{Ca(OH)}_2 \) results in calcium fluoride and water. These reactions are essential in various fields, from industry to bodily functions, as they help control the pH levels. Consider the neutralization of stomach acid by antacids in the human digestive system. Balancing these reactions involves:

• Writing the unbalanced equation with the acid and base forming a salt and water.
• Balancing the salt formed, then adjusting the number of water molecules to match the number of hydrogen atoms available from the acid and hydroxyl groups in the base.
• Ensuring each side of the equation contains equal numbers of each type of atom.

In the case of HF reacting with calcium hydroxide, we double the HF to ensure both fluoride ions and hydroxyl ions are fully utilized, resulting in the balanced equation \( 2 ext{HF} + ext{Ca(OH)}_2 \rightarrow ext{CaF}_2 + 2 ext{H}_2 ext{O} \). This illustrates the principle that no atoms are lost; they are simply rearranged to form new substances.