Fuel formula is given as C15H15S1O3. Calculate the air/fuel ratio for 255 % theoretical air %.

To calculate the air/fuel ratio for 255% theoretical air, we first need to understand the stoichiometric combustion reaction for the given fuel formula C15H15S1O3.

Step 1: Write down the balanced stoichiometric combustion reaction:

C15H15S1O3 + 15O2 + 3.7615N2 -> 15CO2 + 7.5H2O + SO2 + 3.7615N2

This reaction assumes complete combustion where all carbon (C) is converted to carbon dioxide (CO2), all hydrogen (H) to water (H2O), and all sulfur (S) to sulfur dioxide (SO2). Nitrogen (N2) is considered an inert gas and does not react, but it is included in the equation because air is approximately 79% nitrogen and 21% oxygen.

Step 2: Calculate the molar mass of the fuel and air:

Molar mass of C15H15S1O3 = 1512.01 (C) + 151.008 (H) + 32.06 (S) + 3*16.00 (O) = 180.15 + 15.12 + 32.06 + 48.00 = 275.33 g/mol

Molar mass of O2 = 2*16.00 = 32.00 g/mol

Molar mass of N2 = 2*14.01 = 28.02 g/mol

Step 3: Calculate the stoichiometric air/fuel ratio:

Stoichiometric air/fuel ratio = [1532.00 (O2) + 3.7615*28.02 (N2)] / 275.33 (fuel) = 480.00 + 1585.68 / 275.33 = 7.53 kg_air/kg_fuel

Step 4: Calculate the air/fuel ratio for 255% theoretical air:

Air/fuel ratio for 255% theoretical air = 7.53 * 2.55 = 19.20 kg_air/kg_fuel

So, the air/fuel ratio for 255% theoretical air is 19.20 kg_air/kg_fuel.