The correct option is B 101.2 L
2C6H14 +19O2→12CO2 + 14H2O
2 moles of hexane reacts with 19 moles of O2
Molar mass of hexane = 86 g
84×2 g of hexane completely combusts in 19×22.4 L of oxygen at STP.
168 g of hexane completely combusts in 425.6 L of oxygen at STP.
So, the amount of O2 required to completely combust 40 g of hexane =425.6168×40 L of oxygen at STP.
= 101.2 litres of O2