The combustion of ethane is expressed in the balanced reaction C2H6 + 3.5O2= 2CO2 + 3 H2O. Given the mass of oxygen gas, we get the moles of ethane needed by converting this mass to mole (dividing my 32 g/mol), then multiply by 1/3.5 (stioch ratio) and the molar mass of ethane (30 g/mol). The answer is 16.69 grams ethane.