`A` mixture of `C_(3)H_(8)(g) & O_(2)` having total volume `100ml` in an Eudiometry tube is sparked `&` it is observed that a contraction of `45ml` us observed what can be the composition of reacting mixture.

Options

A. `15" " ml" " C_(3)H_(8)" " &" " 85" " ml" " O_(2)`
B. `25" " ml" " C_(3)H_(8)" " & " "75 " "ml" " O_(2)`
C. `45 " "ml" " C_(3)H_(8)" " &" " 55" " ml" " O_(2)`
D. `55 " "ml" " C_(3) H_(8)" " &" " 45 " "ml" " O_(2)`

Explanation

To solve the problem, we need to analyze the reaction between propane (C₃H₈) and oxygen (O₂) and the resulting contraction in volume after the reaction.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The balanced equation for the combustion of propane is:
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]
This indicates that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.

2. **Determine the Total Volume of the Mixture:**
The total volume of the gas mixture is given as 100 ml.

3. **Identify the Contraction in Volume:**
After sparking the mixture, a contraction of 45 ml is observed. This means that the total volume of gases after the reaction is:
\[ \text{Initial Volume} - \text{Contraction} = 100 \text{ ml} - 45 \text{ ml} = 55 \text{ ml} \]

4. **Calculate the Volume of Gases Reacted:**
According to the balanced equation, for every 1 volume of propane (C₃H₈), 5 volumes of oxygen (O₂) are required. Therefore, if we denote the volume of propane as \( V_{C_3H_8} \) and the volume of oxygen as \( V_{O_2} \), we can express the relationship as:
\[ V_{O_2} = 5 \times V_{C_3H_8} \]

5. **Set Up the Volume Equation:**
The total volume of the mixture is:
\[ V_{C_3H_8} + V_{O_2} = 100 \text{ ml} \]
Substituting \( V_{O_2} \):
\[ V_{C_3H_8} + 5 \times V_{C_3H_8} = 100 \text{ ml} \]
\[ 6 \times V_{C_3H_8} = 100 \text{ ml} \]
\[ V_{C_3H_8} = \frac{100}{6} \text{ ml} \approx 16.67 \text{ ml} \]
\[ V_{O_2} = 100 \text{ ml} - V_{C_3H_8} \approx 100 \text{ ml} - 16.67 \text{ ml} \approx 83.33 \text{ ml} \]

6. **Calculate the Volume of Products Formed:**
The volume of carbon dioxide produced can be calculated based on the stoichiometry of the reaction. For every 1 volume of propane, 3 volumes of carbon dioxide are produced:
\[ V_{CO_2} = 3 \times V_{C_3H_8} = 3 \times 16.67 \text{ ml} \approx 50 \text{ ml} \]

7. **Calculate the Final Volume:**
The final volume of gases after the reaction will be:
\[ V_{final} = V_{O_2} - V_{C_3H_8} + V_{CO_2} = 83.33 \text{ ml} - 16.67 \text{ ml} + 50 \text{ ml} = 116.66 \text{ ml} \]
However, we know that the total volume after the reaction is 55 ml, which indicates that not all of the oxygen was consumed.

8. **Possible Compositions:**
Since we have a contraction of 45 ml, we can also consider other possible compositions. If we consider:
- \( V_{C_3H_8} = 15 \text{ ml} \) and \( V_{O_2} = 85 \text{ ml} \)
- \( V_{C_3H_8} = 25 \text{ ml} \) and \( V_{O_2} = 75 \text{ ml} \)

Both of these compositions would yield the same contraction of 45 ml.

### Final Answer:
The possible compositions of the reacting mixture could be:
- 15 ml of propane and 85 ml of oxygen
- 25 ml of propane and 75 ml of oxygen