The compound produced when sodium peroxide is heated with excess sodium is

1. Analyze the question

The question asks for the product formed when sodium peroxide ($\text{Na}_2\text{O}_2$) reacts with excess sodium metal ($\text{Na}$) under heating.

2. Explain the concept

In chemistry, sodium peroxide is a strong oxidizing agent. When it reacts with an excess of a reducing agent like sodium metal, the sodium metal reduces the peroxide to a more stable oxide.

The reaction involves the transfer of electrons from the sodium atoms to the oxygen atoms within the peroxide structure. Because there is an excess of sodium, the reaction proceeds until the sodium peroxide is fully converted into the most stable form of sodium oxide, which is sodium oxide ($\text{Na}_2\text{O}$).

3. Show the solution

The chemical equation for this reaction is:

• Reactants: Sodium peroxide ($\text{Na}_2\text{O}_2$) and Sodium ($\text{Na}$).
• Process: The sodium metal reacts with the oxygen in the peroxide. Since the sodium is in excess, all the peroxide is consumed to form the simple oxide.
• Product: Sodium oxide ($\text{Na}_2\text{O}$).

4. Identify the answer

Comparing our result to the given options: A. $\text{Na}_2\text{CO}_3$ (Sodium trioxocarbonate IV) B. $\text{Na}_2\text{O}$ (Sodium oxide) C. $\text{NaNO}_2$ (Sodium trioxonitrate III) D. $\text{NaOH}$ (Sodium hydroxide)

The correct option is B.

5. Tip

When dealing with Group 1 metals (like Sodium) and oxygen, remember the "excess" rule:

• If there is excess oxygen, you typically get the peroxide ($\text{Na}_2\text{O}_2$).
• If there is excess metal, you typically get the simple oxide ($\text{Na}_2\text{O}$).