When steam is passed over red hot carbon the substances produced are

Analysis

The question asks for the chemical products formed when steam (water vapor) reacts with red-hot carbon. This is a classic industrial process known as the production of water gas.

Concept

When steam ($H_2O$) is passed over red-hot carbon (coke) at a high temperature (approximately 1000°C), a redox reaction occurs. The carbon acts as a reducing agent, removing oxygen from the water molecule.

The chemical equation for this reaction is: $C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)$

• $C$: Carbon (red-hot coke)
• $H_2O$: Steam
• $CO$: Carbon(II) oxide (also known as carbon monoxide)
• $H_2$: Hydrogen gas

The mixture of carbon(II) oxide and hydrogen is collectively called water gas, which is an important industrial fuel.

Step-by-Step Solution

1. Identify the reactants: Carbon ($C$) and Steam ($H_2O$).
2. Determine the reaction type: This is a displacement reaction where carbon reduces steam.
3. Balance the equation: • One atom of Carbon on the left, one on the right. • Two atoms of Hydrogen on the left, two on the right. • One atom of Oxygen on the left, one on the right. • The equation $C + H_2O \rightarrow CO + H_2$ is balanced.
4. Identify the products: The products are Carbon(II) oxide ($CO$) and Hydrogen gas ($H_2$).

Comparing this to your options:

• Option A mentions trioxocarbonate(IV) acid (carbonic acid), which is incorrect.
• Option B mentions oxygen, which is not produced.
• Option C matches our derived products: hydrogen and carbon(II) oxide.
• Option D mentions carbon(IV) oxide ($CO_2$), which is only produced if the reaction occurs at a lower temperature or with excess steam, but the standard reaction for "red hot carbon" is the production of water gas.

Exam Tip

Remember the term "Water Gas". Whenever you see "steam + red hot carbon," immediately think of the mixture $CO + H_2$. If the question had said "excess steam," then carbon(IV) oxide ($CO_2$) would be produced instead of carbon(II) oxide, but for standard red-hot carbon, stick with $CO$.