The equvivalent weight of `CH_(4)` in the reaction `CH_(4)+2O_(2)to CO_(2)+2H_(2)O` is [M=mol. Wt]

Options

A. M/4
B. M/8
C. M/12
D. M/16

Explanation

To find the equivalent weight of \( CH_4 \) in the reaction

\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O, \]

we will follow these steps:

### Step 1: Understand the concept of equivalent weight
The equivalent weight (EW) of a substance can be calculated using the formula:

\[ EW = \frac{M}{n} \]

where \( M \) is the molecular weight of the substance and \( n \) is the number of equivalents (n-factor).

### Step 2: Determine the molecular weight of \( CH_4 \)
The molecular weight of \( CH_4 \) (methane) is calculated as follows:
- Carbon (C) has a molar mass of approximately 12 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.

Thus, the molecular weight of \( CH_4 \) is:

\[ M = 12 + (4 \times 1) = 12 + 4 = 16 \, \text{g/mol}. \]

### Step 3: Determine the n-factor for \( CH_4 \)
The n-factor is determined by the change in oxidation state of the element of interest during the reaction.

- In \( CH_4 \), the oxidation state of carbon is \( -4 \) (since hydrogen has an oxidation state of \( +1 \)).
- In \( CO_2 \), the oxidation state of carbon is \( +4 \).

The change in oxidation state for carbon is:

\[ \text{Change} = +4 - (-4) = +4 + 4 = 8. \]

Thus, the n-factor for \( CH_4 \) in this reaction is \( 8 \).

### Step 4: Calculate the equivalent weight of \( CH_4 \)
Now that we have both the molecular weight and the n-factor, we can substitute these values into the equivalent weight formula:

\[ EW = \frac{M}{n} = \frac{16 \, \text{g/mol}}{8} = 2 \, \text{g/equiv}. \]

### Conclusion
The equivalent weight of \( CH_4 \) in the given reaction is \( 2 \, \text{g/equiv} \).

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