The volume of `O_(2)` at STP required for the complete combustion of 4g `CH_(4)` is

Options

A. 5.6 litre
B. 2.88 litre
C. 22.4 litre
D. 11.2 litre

Answer

To find the volume of O₂ at STP required for the complete combustion of 4 g of CH₄ (methane), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane.
The balanced equation for the complete combustion of methane is:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Determine the molar mass of CH₄.
The molar mass of CH₄ (methane) can be calculated as follows:
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol
Thus, the molar mass of CH₄ is:
\[ 12 + 4 = 16 \text{ g/mol} \]

### Step 3: Calculate the number of moles of CH₄ in 4 g.
Using the molar mass, we can calculate the number of moles of CH₄ in 4 g:
\[ \text{Number of moles of CH}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{16 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 4: Use the stoichiometry of the reaction to find the moles of O₂ required.
From the balanced equation, we see that 1 mole of CH₄ reacts with 2 moles of O₂. Therefore, for 0.25 moles of CH₄, the moles of O₂ required will be:
\[ \text{Moles of O}_2 = 0.25 \text{ moles CH}_4 \times 2 = 0.5 \text{ moles O}_2 \]

### Step 5: Calculate the volume of O₂ at STP.
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the volume of 0.5 moles of O₂ is:
\[ \text{Volume of O}_2 = 0.5 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ L} \]

### Final Answer:
The volume of O₂ required for the complete combustion of 4 g of CH₄ is **11.2 liters**.

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