Question

One mole of Propane (C₃H₈) on combustion given 'x' kilo joules at STP. Heat liberated by the combustion of 11.2 litres of Propane at STP in kilo joules is

• A. \(x\)
• B. \(\frac x2\)
• C. \(11.2 x\)
• D. \(2 x\)

Answer 1

The Correct Option is B

To solve the problem, we need to calculate the heat liberated by the combustion of 11.2 liters of propane (C₃H₈) at STP, given that the heat released by one mole of propane is \(x\) kilojoules.

1. Understanding the Relationship:
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. So, the volume of 1 mole of propane (C₃H₈) is 22.4 liters at STP.

2. Given:
- Heat released by 1 mole of propane = \(x\) kilojoules.
- Volume of 1 mole of propane at STP = 22.4 liters.
- Volume of propane given = 11.2 liters.

3. Using the Proportionality:
The heat released is proportional to the volume of propane. Since 22.4 liters of propane release \(x\) kilojoules of heat, 11.2 liters (which is half the volume of 22.4 liters) will release half the heat. Therefore, the heat released by the combustion of 11.2 liters of propane is:
\[ \frac{x}{2} \]

Final Answer:
The heat liberated by the combustion of 11.2 liters of propane at STP is \( \mathbf{\frac{x}{2}} \) kilojoules.