Right answer is (a) 1/23 [8/3 C + 8(H – (O/8)) + S]
Explanation: Let C, H, O and S represent percentage by mass of carbon(C), Hydrogen (H2), oxygen and sulfur respectively.
The mass of oxygen required for complete combustion of fuel is given by,
= 1/100 [8/3 C + 8H – O + S]
= 1/100 [8/3C + 8(H – (0/8)) + S]
As air contains 23% of oxygen by mass, minimum air required for burning one kg of liquid fuel completely is given by,
Min. air required = 1/100 [8/3 C + 8(H – (O/8)) + S] 100/23
= 1/23 [8/3 C + 8 (H – (O/8)) + S].