When a certain amount of ethylene was combusted, 6226 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ, the volume of `O_(2)` (at STP) that entered into the reaction is :

Options

A. 296.5 mL
B. 296.5 litre
C. `6226 xx 22.4` litre
D. 22.4 litre

Explanation

To solve the problem of determining the volume of \( O_2 \) that entered into the reaction when a certain amount of ethylene was combusted, we can follow these steps:

### Step 1: Understand the Heat of Combustion
The heat of combustion of ethylene (C₂H₄) is given as 1411 kJ. This means that when 1 mole of ethylene is combusted, it releases 1411 kJ of heat.

### Step 2: Calculate the Number of Moles of Ethylene
We know that 6226 kJ of heat was evolved during the combustion. To find the number of moles of ethylene that were combusted, we can use the formula:

\[ \text{Number of moles of C₂H₄} = \frac{\text{Total heat evolved}}{\text{Heat of combustion per mole}} \]

Substituting the values:

\[ \text{Number of moles of C₂H₄} = \frac{6226 \text{ kJ}}{1411 \text{ kJ/mole}} \approx 4.41 \text{ moles} \]

### Step 3: Determine the Moles of Oxygen Required
From the balanced chemical equation for the combustion of ethylene:

\[ C₂H₄ + 3O₂ \rightarrow 2CO₂ + 2H₂O \]

We see that 1 mole of ethylene reacts with 3 moles of oxygen. Therefore, the number of moles of \( O_2 \) required for 4.41 moles of ethylene is:

\[ \text{Number of moles of } O₂ = 4.41 \text{ moles of C₂H₄} \times 3 = 13.23 \text{ moles of } O₂ \]

### Step 4: Calculate the Volume of Oxygen at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 13.23 moles of \( O_2 \) can be calculated as:

\[ \text{Volume of } O₂ = \text{Number of moles of } O₂ \times 22.4 \text{ L/mole} \]

Substituting the values:

\[ \text{Volume of } O₂ = 13.23 \text{ moles} \times 22.4 \text{ L/mole} \approx 296.35 \text{ liters} \]

### Conclusion
Thus, the volume of \( O_2 \) that entered into the reaction is approximately **296.35 liters**.

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