Question

A hydrocarbon \( C_n H_m \) is burnt in air (O\(_2\) + 3.78N\(_2\)). The stoichiometric fuel to air mass ratio for this process is 
Note: Atomic Weight: C(12), H(1) Effective Molecular Weight: Air(28.8)

• A. \( 0.0291 \left( \frac{4n + m}{12n + m} \right) \)
• B. \( 34.42 \left( \frac{12n + m}{4n + m} \right) \)
• C. \( 34.42 \left( \frac{4n + m}{12n + m} \right) \)
• D. \( 0.0291 \left( \frac{12n + m}{4n + m} \right) \)

Hint:

In combustion reactions, balancing the fuel and air components requires attention to atomic weights and stoichiometric relationships. Ensure that the mass of oxygen and air are correctly calculated to determine the correct ratio.

Answer 1

The Correct Option is D

For the combustion of a hydrocarbon \( C_n H_m \) in air (with \( O_2 + 3.78N_2 \)), the stoichiometric reaction is: \[ C_n H_m + aO_2 + bN_2 \rightarrow xCO_2 + yH_2O, \] where \( a \) and \( b \) are determined based on the molecular composition of the fuel. For carbon: \( c : n = x \), so \( x = n \).
For hydrogen: \( H : m = 2y \), so \( y = \frac{m}{2} \).
For oxygen: \( O : 2a = 2x + y \), so \( a = \frac{2n + m}{4} \).
Mass of \( O_2 \) and mass of air can be calculated as follows:
\[ {Mass of } O_2 = 32 \times \left( \frac{n + m}{4} \right), \] \[ {Mass of air} = 137.84 \times \left( \frac{4n + m}{4} \right), \] Finally, the fuel to air ratio is: \[ {Fuel to Air Ratio} = \frac{12n + m}{4n + m} = 0.0291 \left( \frac{12n + m}{4n + m} \right). \] Thus, the correct answer is option (D).