When a certain amount of ethylene was combusted, `5644 kJ` heat was evolved. If heat of combustion of ethylene is `1411 kJ`, the volume of `O_(2)` (at `NTP`) the entered into the reaction is:

Options

A. `268.8 ml`
B. `268.8 L`
C. `6226xx22.4 L`
D. `22.4 L`

Answer

To solve the problem, we need to determine the volume of oxygen (O₂) that reacted with ethylene (C₂H₄) during combustion, given the heat evolved and the heat of combustion of ethylene.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The combustion of ethylene can be represented by the following balanced chemical equation:
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]
From this equation, we see that 1 mole of ethylene reacts with 3 moles of oxygen.

2. **Determine the Moles of Ethylene Combusted**:
The heat evolved during the combustion of ethylene is given as \( 5644 \, \text{kJ} \). The heat of combustion of ethylene is given as \( 1411 \, \text{kJ/mol} \). We can find the number of moles of ethylene that were combusted using the formula:
\[ \text{Moles of } C_2H_4 = \frac{\text{Heat evolved}}{\text{Heat of combustion per mole}} = \frac{5644 \, \text{kJ}}{1411 \, \text{kJ/mol}} \]
Calculating this gives:
\[ \text{Moles of } C_2H_4 = 4 \, \text{moles} \]

3. **Calculate the Moles of Oxygen Used**:
From the balanced equation, we know that 1 mole of C₂H₄ reacts with 3 moles of O₂. Therefore, the moles of O₂ consumed can be calculated as:
\[ \text{Moles of } O_2 = 3 \times \text{Moles of } C_2H_4 = 3 \times 4 = 12 \, \text{moles} \]

4. **Convert Moles of Oxygen to Volume**:
At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies \( 22.4 \, \text{L} \). Thus, the volume of O₂ can be calculated as:
\[ \text{Volume of } O_2 = \text{Moles of } O_2 \times 22.4 \, \text{L/mol} = 12 \, \text{moles} \times 22.4 \, \text{L/mol} \]
Calculating this gives:
\[ \text{Volume of } O_2 = 268.8 \, \text{L} \]

### Final Answer:
The volume of O₂ that entered into the reaction is \( 268.8 \, \text{L} \).

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