Question

4\. Complete the following equations: (i) \(\left.\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CW}\right)+\mathrm{H}_{2} \mathrm{O}_{2}\) (ii) \(2 \mathrm{KO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \underline{(X)}+\underline{(Y)}+\mathrm{O}_{2}\) (iii) \(\mathrm{Na}_{2} \mathrm{O}+\mathrm{CO}_{2} \rightarrow \underline{(Z)}\) \(\begin{array}{llll}W & X & Y & Z\end{array}\) \(\begin{array}{lllll}\text { (a) } 4 \mathrm{Na} & \mathrm{K}_{2} \mathrm{O} & \mathrm{H}_{2} \mathrm{O} & \mathrm{Na}_{2} \mathrm{O}_{2}\end{array}\) \(\begin{array}{lllll}\text { (b) } 4 \mathrm{Na} & \mathrm{K}_{2} \mathrm{O} & \mathrm{H}_{2} \mathrm{O}_{2} & \mathrm{Na}_{2} \mathrm{CO}_{3}\end{array}\) (c) \(4 \mathrm{NaOH} \quad 2 \mathrm{KOH} \quad \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Na}_{2} \mathrm{O}_{2}\) \(\begin{array}{lllll}\text { (d) } 2 \mathrm{NaOH} & 2 \mathrm{KOH} & \mathrm{H}_{2} \mathrm{O}_{2} & \mathrm{Na}_{2} \mathrm{CO}_{3}\end{array}\)

Short answer

W is 2 \(\mathrm{NaOH}\) (sodium hydroxide), X is 2 \(\mathrm{KOH}\) (potassium hydroxide), Y is \(\mathrm{H}_2\mathrm{O}_2\) (hydrogen peroxide), and Z is \(\mathrm{Na}_2\mathrm{CO}_3\) (sodium carbonate). The corresponding options are (d) for W, (c) for X, (d) for Y, and (d) for Z.

Step-by-step solution

- Solve Equation (i)

To balance the reaction of \(\mathrm{Na}_2\mathrm{O}_2\) with \(\mathrm{H}_2\mathrm{O}\), we need to determine what product CW is. Considering the reactants, the sodium in \(\mathrm{Na}_2\mathrm{O}_2\) will likely form a hydroxide with water. The balanced equation should be \(\mathrm{Na}_2\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{NaOH} + \mathrm{H}_2\mathrm{O}_2\). Therefore, W is likely to be \(\mathrm{NaOH}\).

- Solve Equation (ii)

Given the reactants \(2\mathrm{KO}_2\) and \(2\mathrm{H}_2\mathrm{O}\), and knowing that one of the products is \(\mathrm{O}_2\), it is likely that potassium hydroxide (\(\mathrm{KOH}\)) will be produced from \(\mathrm{KO}_2\) and water. We also know that the subscript for X must be 2, since there are 2 potassium atoms in the reactants. The balanced equation will be \(2\mathrm{KO}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{KOH} + \mathrm{H}_2\mathrm{O}_2 + \mathrm{O}_2\). So, X is \(\mathrm{KOH}\) and Y is \(\mathrm{H}_2\mathrm{O}_2\).

- Solve Equation (iii)

When \(\mathrm{Na}_2\mathrm{O}\) reacts with \(\mathrm{CO}_2\), the expected product is a carbonate. Combining sodium oxide with carbon dioxide will form sodium carbonate. Thus, the balanced equation is \(\mathrm{Na}_2\mathrm{O} + \mathrm{CO}_2 \rightarrow \mathrm{Na}_2\mathrm{CO}_3\). Therefore, Z is \(\mathrm{Na}_2\mathrm{CO}_3\).

Key concepts

Chemical Reaction Balancing

Balancing chemical equations is a fundamental skill in chemistry, one that ensures the law of conservation of mass is upheld. The principle behind this is straightforward: the number of atoms for each element must be the same on both the reactant and product sides of an equation.

For students grappling with this concept, it's akin to solving a puzzle – each piece, or in this case, each atom and molecule, must fit perfectly to complete the picture. Starting with the more complex molecules often simplifies this process. For instance, balance elements that appear in a single reactant and product first, followed by those in multiple substances, and save hydrogen and oxygen for last as they're usually found in multiple compounds.

Sodium Peroxide Reactions

Sodium peroxide (\textbf{Na}\(_2\)\textbf{O}\(_2\)) often acts as a powerful oxidizing agent, reacting exothermically with water to produce sodium hydroxide (\textbf{NaOH}) and hydrogen peroxide (\textbf{H}\(_2\)\textbf{O}\(_2\)). This reaction is an excellent illustration of redox processes and reactivity in inorganic chemistry.

Understanding Sodium Peroxide Reactivity

• Sodium peroxide donates oxygen to water, illustrating its oxidizing properties.
• It's important to recognize the stoichiometry of the reaction, typically requiring two water molecules for every molecule of sodium peroxide to balance.
• Students should take note of the state symbols (s, l, g, aq) which provide clues to the conditions under which these reactions occur.

Recognizing patterns in reactions like these can help students predict products in similar scenarios.

Potassium Superoxide Reactions

Potassium superoxide (\textbf{KO}\(_2\)) has its unique chemistry, different from that of sodium peroxide despite the similarities in their names. In the presence of water, potassium superoxide releases oxygen gas and forms potassium hydroxide (\textbf{KOH}) along with hydrogen peroxide (\textbf{H}\(_2\)\textbf{O}\(_2\)).

This is particularly illustrative for students learning about peroxides and superoxides, as it demonstrates how these substances can act as a source of oxygen – a concept critical in various fields such as respiratory technologies and space travel where oxygen generation is required.

Paying Attention to Oxidation States

Understanding the changes in oxidation states during these reactions is important for deeper insights. It also underscores the utility of balancing reactions to ensure that the changes in oxidation numbers match the electron transfer taking place.