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Step-by-Step Solution
Step 1: Write the General Combustion Reaction
For an alkane with the molecular formula $C_xH_y$, the complete combustion reaction is given by:
$ C_xH_y + \bigl(x + \tfrac{y}{4}\bigr) O_2 \;\to\; x\,CO_2 \;+\;\tfrac{y}{2}\,H_2O $
Step 2: Relate Volume to Number of Moles (Avogadro's Law)
At the same temperature and pressure, equal volumes of gases contain equal numbers of moles. Therefore, the volume of a gas is directly proportional to the number of moles.
Thus, if $C_xH_y$ and $O_2$ are both measured under the same conditions of temperature and pressure, the volume ratio will reflect the mole ratio directly.
Step 3: Set Up the Volume Ratio Equation
From the combustion reaction, the ratio of moles of $O_2$ needed per mole of $C_xH_y$ is $\bigl(x + \tfrac{y}{4}\bigr)$. By Avogadro’s law, the same ratio applies to their volumes:
$\text{Required volume of }O_2 = \bigl(x + \tfrac{y}{4}\bigr) \times \text{(volume of }C_xH_y\text{)}$
Given in the problem:
• 5 L of $C_xH_y$ require 25 L of $O_2$.
So we write:
$5\bigl(x + \tfrac{y}{4}\bigr) = 25$
Simplifying:
$x + \tfrac{y}{4} = 5$
Step 4: Check Each Option
Ethane ($C_2H_6$):
$x = 2, \quad y = 6$
$x + \tfrac{y}{4} = 2 + \tfrac{6}{4} = 2 + 1.5 = 3.5 \ne 5$
Does not match the required value of 5.
Propane ($C_3H_8$):
$x = 3, \quad y = 8$
$x + \tfrac{y}{4} = 3 + \tfrac{8}{4} = 3 + 2 = 5$
This matches the required value of 5.
Butane ($C_4H_{10}$) and Isobutane ($C_4H_{10}$):
$x = 4, \quad y = 10$
$x + \tfrac{y}{4} = 4 + \tfrac{10}{4} = 4 + 2.5 = 6.5 \ne 5$
Also does not match the required value of 5.
Step 5: Conclusion
Since $x + \tfrac{y}{4} = 5$ only for $C_3H_8$, the alkane that requires 25 L of $O_2$ to combust 5 L of the alkane is Propane.
