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?
Volume of `N_(2)` present in the mixture?

Options

A. `2ml`
B. `4ml`
C. `6ml`
D. `8ml`

Explanation

To solve the problem step by step, let's break down the information given and analyze it systematically.

### Step 1: Understanding the Initial Mixture
We have a 10 ml mixture of three gases: nitrogen (N₂), an alkane (CnH₂n), and oxygen (O₂). When this mixture undergoes combustion, we observe a contraction of 2 ml. This means that the total volume of gases after combustion is 10 ml - 2 ml = 8 ml.

### Step 2: Residual Gases After Combustion
After the combustion, the residual gases are passed through KOH, which absorbs carbon dioxide (CO₂). The contraction of 2 ml indicates that 2 ml of gas was consumed during the combustion process, which is primarily due to the formation of CO₂ and water vapor.

### Step 3: Analyzing the Remaining Gases
After passing through KOH, the remaining gases consist of nitrogen (N₂) and any unreacted oxygen (O₂). Since KOH absorbs CO₂, we can conclude that the remaining gases are N₂ and O₂.

### Step 4: Adding Excess Hydrogen (H₂)
Next, we add excess hydrogen gas (H₂) to the remaining mixture. The reaction that occurs is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
This indicates that nitrogen reacts with hydrogen to form ammonia (NH₃).

### Step 5: Volume Change After Adding H₂
After the addition of H₂ and subsequent combustion, the volume of gas is absorbed by water, causing a reduction of 8 ml. This means that the total volume of gas after the reaction is reduced by 8 ml due to the formation of ammonia.

### Step 6: Calculating the Volume of Ammonia Produced
From the reaction stoichiometry, we know that:
- 1 mole of N₂ produces 2 moles of NH₃.
- The volume of gases is proportional to the number of moles (at the same temperature and pressure).

Since the reduction in volume is 8 ml, this corresponds to the volume of ammonia produced. Therefore, the volume of ammonia produced is 8 ml.

### Step 7: Relating Ammonia Volume to Nitrogen Volume
From the stoichiometry of the reaction:
- 2 volumes of NH₃ are produced from 1 volume of N₂.
- Therefore, if 8 ml of NH₃ is produced, the volume of N₂ used is:
\[ \text{Volume of } N_2 = \frac{8 \text{ ml}}{2} = 4 \text{ ml} \]

### Final Answers
1. **Gas produced after introduction of H₂**: Ammonia (NH₃).
2. **Volume of N₂ present in the mixture**: 4 ml.