If `v mL` of a gaseous hydrocarbon, after explosion with excess of oxygen, showed a contraction of `2.5 v mL ` and a further contraction of `2v mL` with caustic potash, Find the formula of hydrocarbon.

Explanation

To solve the problem, we need to analyze the information given about the gaseous hydrocarbon and the reactions it undergoes. Let's break it down step by step.

### Step 1: Understanding the Reaction
The gaseous hydrocarbon can be represented as \( C_nH_{2n+2} \) (which is the general formula for alkanes). When it combusts with excess oxygen, it produces carbon dioxide and water:
\[ C_nH_{2n+2} + \left(3n + \frac{1}{2}\right) O_2 \rightarrow n CO_2 + \left(n + 1\right) H_2O \]

### Step 2: Analyzing the Volume Contraction
1. The initial volume of the hydrocarbon is \( v \) mL.
2. After combustion, the volume contracts by \( 2.5v \) mL. This contraction is due to the formation of water vapor. Since water condenses, it is no longer in the gaseous state, leading to a reduction in volume.
3. The remaining volume after the first contraction is:
\[ v - 2.5v = -1.5v \text{ (which indicates a mistake in interpretation; we should consider the volume of gases remaining)} \]

### Step 3: Relating Contraction to Moles
The first contraction of \( 2.5v \) mL indicates that the volume of water produced is equivalent to this contraction. The number of moles of water produced is \( n + 1 \) (from the balanced equation).

Since the volume of gas is proportional to the number of moles, we can say:
\[ n + 1 = 2.5 \]
From this, we can solve for \( n \):
\[ n = 2.5 - 1 = 1.5 \]

### Step 4: Considering the Second Contraction
The second contraction of \( 2v \) mL is due to the carbon dioxide produced. The number of moles of carbon dioxide produced is \( n \) (from the balanced equation).

Thus, we can write:
\[ n = 2 \]

### Step 5: Finding the Hydrocarbon Formula
From the value of \( n \):
- If \( n = 2 \), then the hydrocarbon is \( C_2H_{2(2)+2} = C_2H_6 \).

### Conclusion
The formula of the hydrocarbon is \( C_2H_6 \) (ethane).

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