How much volume of oxygen at STP in litres is required to burn 4g of methane gas completely?

Options

A. 11.2
B. 5.6
C. 2.8
D. 8

Answer

To find the volume of oxygen required to burn 4 grams of methane gas completely at STP (Standard Temperature and Pressure), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane.
The combustion of methane (CH₄) in the presence of oxygen (O₂) can be represented by the following balanced equation:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of methane.
The molar mass of methane (CH₄) is calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
- Total molar mass of CH₄ = 12 g/mol + 4 g/mol = 16 g/mol

### Step 3: Determine the amount of oxygen required for the combustion of methane.
From the balanced equation, we see that 1 mole of methane requires 2 moles of oxygen for complete combustion.

### Step 4: Calculate the volume of oxygen required for 16 grams of methane.
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, for 16 grams of methane (which is 1 mole), the volume of oxygen required is:
\[ 2 \text{ moles of O}_2 \times 22.4 \text{ L/mole} = 44.8 \text{ L} \]

### Step 5: Calculate the volume of oxygen required for 4 grams of methane.
Since 4 grams of methane is \(\frac{4 \text{ g}}{16 \text{ g/mol}} = 0.25 \text{ moles}\), we can find the volume of oxygen needed:
1. For 1 mole of methane, 2 moles of oxygen are required.
2. Therefore, for 0.25 moles of methane, the moles of oxygen required will be:
\[ 2 \times 0.25 = 0.5 \text{ moles of O}_2 \]

3. Now, we can calculate the volume of oxygen required:
\[ 0.5 \text{ moles of O}_2 \times 22.4 \text{ L/mole} = 11.2 \text{ L} \]

### Final Answer:
The volume of oxygen required to burn 4 grams of methane completely at STP is **11.2 liters**.

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