Question

The general combustion equation for any alkane is given by the following equation: \(C_nH_{2n+2} + X O_2 \to nCO_2 + (n+1) H_2O\). What is the value of ‘X’ in the above equation?

• A. \((n+1)/2\)
• B. \((2n+1)/2\)
• C. \((2n+2)/2\)
• D. \((3n+1)/2\)
• E. \((4n+1)/2\)

Hint:

For any hydrocarbon \(C_xH_y\), the balanced oxygen coefficient is always \(x + y/4\). For alkanes, \(x=n\) and \(y=2n+2\). So, \(n + (2n+2)/4 = n + (n+1)/2 = (3n+1)/2\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Combustion is the reaction of a hydrocarbon with oxygen to produce carbon dioxide and water. To find the coefficient of oxygen (\(X\)), we must balance the oxygen atoms on both sides of the equation.

Step 2: Key Formula or Approach:
The total number of oxygen atoms on the product side must equal the total number of oxygen atoms on the reactant side.

Step 3: Detailed Explanation:
Given Equation: \(C_nH_{2n+2} + X O_2 \to nCO_2 + (n+1) H_2O\)
1. Count Oxygen atoms in products:
- From \(nCO_2\): \(2 \times n = 2n\) atoms.
- From \((n+1)H_2O\): \(1 \times (n+1) = n+1\) atoms.
- Total Oxygen atoms in products = \(2n + n + 1 = 3n + 1\).
2. Count Oxygen atoms in reactants:
- From \(X O_2\): \(2 \times X = 2X\) atoms.
3. Equating both sides:
\[ 2X = 3n + 1 \]
\[ X = \frac{3n + 1}{2} \]

Step 4: Final Answer:
The value of \(X\) is \((3n+1)/2\).