`CH_(4(g))+2O_(2(g))rarr CO_(2(g))+H_(2)O_((l))`, as per this reaction, if 20 ml of `CH_(4)` and 20 ml of `O_(2)` were exploded together than the reaction mixture will have a volume of

Options

A. 40 ml
B. 20 ml
C. 60 ml
D. 30 ml

Explanation

To solve the problem, we need to analyze the reaction and the volumes of the gases involved. The reaction given is:

\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

### Step 1: Identify the stoichiometric coefficients
From the balanced equation, we see that:
- 1 volume of CH₄ reacts with 2 volumes of O₂ to produce 1 volume of CO₂ and 2 volumes of H₂O (which is a liquid and does not contribute to the gas volume).

### Step 2: Determine the limiting reagent
We are given:
- 20 ml of CH₄
- 20 ml of O₂

According to the stoichiometry:
- 1 ml of CH₄ requires 2 ml of O₂.
- Therefore, 20 ml of CH₄ would require \( 20 \times 2 = 40 \) ml of O₂.

Since we only have 20 ml of O₂ available, O₂ is the limiting reagent.

### Step 3: Calculate the amount of CH₄ that reacts
Since O₂ is the limiting reagent, we need to find out how much CH₄ reacts with the available O₂:
- From the stoichiometry, 2 ml of O₂ reacts with 1 ml of CH₄.
- Therefore, 20 ml of O₂ will react with \( \frac{20}{2} = 10 \) ml of CH₄.

### Step 4: Calculate the remaining CH₄
Initially, we had 20 ml of CH₄, and 10 ml of it reacted with O₂. Thus, the remaining CH₄ is:
\[ 20 \text{ ml} - 10 \text{ ml} = 10 \text{ ml} \]

### Step 5: Calculate the volume of the reaction mixture
After the reaction, we have:
- 10 ml of unreacted CH₄
- 20 ml of O₂ (all of it was consumed)
- 10 ml of CO₂ produced (since 20 ml of O₂ produces 10 ml of CO₂).

Now, we can calculate the total volume of the reaction mixture:
\[ \text{Total volume} = \text{Volume of unreacted CH}_4 + \text{Volume of CO}_2 \]
\[ \text{Total volume} = 10 \text{ ml (CH}_4) + 10 \text{ ml (CO}_2) = 20 \text{ ml} \]

### Final Answer
The volume of the reaction mixture after the explosion is **20 ml**.

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