When 80 gm `CH_(4)` is burnt completely , CO and `CO_(2)` gases are formed in 1:4 mole ratio. What is the mass of `O_(2)` gas used in combustion.

Answer

To solve the problem of determining the mass of \( O_2 \) gas used in the combustion of 80 grams of methane (\( CH_4 \)), we will follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane.
The combustion of methane can be represented by the following balanced equation:
\[ CH_4 + 2O_2 \rightarrow CO + 2H_2O + 4CO_2 \]
However, since the question states that the products are \( CO \) and \( CO_2 \) in a 1:4 mole ratio, we can adjust the equation accordingly:
\[ CH_4 + 2O_2 \rightarrow CO + 2H_2O + 4CO_2 \]

### Step 2: Calculate the number of moles of \( CH_4 \).
To find the number of moles of methane, we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
The molar mass of \( CH_4 \) (methane) is calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol × 4 = 4 g/mol
- Total molar mass of \( CH_4 \) = 12 + 4 = 16 g/mol

Now, substituting the values:
\[ \text{Number of moles of } CH_4 = \frac{80 \text{ g}}{16 \text{ g/mol}} = 5 \text{ moles} \]

### Step 3: Determine the number of moles of \( O_2 \) required.
From the balanced equation, we see that 1 mole of \( CH_4 \) reacts with 2 moles of \( O_2 \). Therefore, for 5 moles of \( CH_4 \):
\[ \text{Moles of } O_2 = 5 \text{ moles } CH_4 \times 2 \text{ moles } O_2/1 \text{ mole } CH_4 = 10 \text{ moles } O_2 \]

### Step 4: Calculate the mass of \( O_2 \).
To find the mass of \( O_2 \), we again use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
The molar mass of \( O_2 \) is:
- Oxygen (O) = 16 g/mol × 2 = 32 g/mol

Now substituting the values:
\[ \text{Mass of } O_2 = 10 \text{ moles} \times 32 \text{ g/mol} = 320 \text{ g} \]

### Final Answer:
The mass of \( O_2 \) gas used in the combustion is **320 grams**.

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