Comprehension # 9
A `10ml` mixture of `N_(2)`, a alkane `& O_(2)` undergo combustion in Eudiometry tube. There was contraction of `2ml`, when residual gases are passed are passed through `KOH`. To the remaining mixture comprising of only one gas excess `H_(2)` was added `&` after combustion the gas product is absorbed by water. causing a reduction in volume of `8ml`.
Gas produced after introduction of `H_(2)` in the mixture?

Options

A. `H_(2)O`
B. `CH_(4)`
C. `CO_(2)`
D. `NH_(3)`

Explanation

To solve the problem step by step, we need to analyze the combustion process and the gases involved.

### Step 1: Understand the Initial Mixture
The initial mixture consists of:
- Nitrogen gas (N₂)
- An alkane (CnH2n+2)
- Oxygen gas (O₂)

The total volume of the mixture is 10 ml.

### Step 2: Analyze the Combustion Reaction
When the mixture undergoes combustion, the alkane reacts with oxygen to produce carbon dioxide (CO₂) and water (H₂O). The general reaction for the combustion of an alkane can be written as:
\[ C_nH_{2n+2} + O_2 \rightarrow CO_2 + H_2O \]

### Step 3: Determine the Volume Contraction
The problem states that there was a contraction of 2 ml after combustion. This means that 2 ml of gases were consumed during the reaction. Since the alkane reacts with oxygen, we can infer that the volume of CO₂ produced is equal to the volume of O₂ consumed.

### Step 4: Calculate the Residual Gases
After the combustion, the residual gases consist of:
- Unreacted nitrogen (N₂)
- Any excess oxygen (if present)

Since the contraction was 2 ml, the remaining volume of gases after combustion is:
\[ 10 \, \text{ml} - 2 \, \text{ml} = 8 \, \text{ml} \]

### Step 5: Passing Residual Gases through KOH
The residual gases are passed through potassium hydroxide (KOH), which absorbs carbon dioxide (CO₂). The problem states that after passing through KOH, there is a remaining volume of gases. This indicates that the 2 ml of CO₂ produced during combustion was absorbed, leaving only nitrogen and any excess oxygen.

### Step 6: Add Excess Hydrogen
The remaining mixture now comprises only one gas (either N₂ or excess O₂). When excess hydrogen (H₂) is added, it reacts with nitrogen (N₂) to form ammonia (NH₃):
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 7: Analyze the Volume Reduction
After the reaction with hydrogen, the gas product is absorbed by water, causing a reduction in volume of 8 ml. This indicates that the total volume of gas produced after the reaction is equal to the volume of ammonia formed.

### Step 8: Conclusion
The gas produced after the introduction of H₂ in the mixture is ammonia (NH₃).

### Final Answer
The gas produced after the introduction of H₂ in the mixture is **NH₃**.

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