This reaction shows the complete combustion of octane, c8h18, a component of gasoline. 2 c8h18(l) + 25 o2(g) 16 co2(g) + 18 h2o(l) (a) how many moles of o2 are needed to burn 1.60 mol of c8h18? mol (b) how many moles of co2 are produced when 0.19 mol of c8h18 are burned? mol (c) how many grams of o2 are needed to burn 1.15 g of c8h18? g

The balanced combustion reaction of octane ( which is a component of gasoline) can be written as:

C₈H₈ + O₂→ 8 CO₂ + 9 H₂O.

(a) For combustion of one mole of butane, moles of oxygen is required. So, for combustion of 1.60 mole of butane, Oxygen required is mole= 20 moles.

(b) On combustion of one mole of butane 8 moles of carbon dioxide (CO₂) is produced. Hence when 0.19 moles of butane is burnt, carbon dioxide produced is= 0.19 X 8 moles= 1.52 moles.

(c) Molar mass of butane is= (8 X12)+ (8 X 1)= 104 g/mol and molar mass of

O₂ is 32 g/mol.

As per balanced combustion reaction, to burn one mole butane or 104 g butane (25/2) mole of oxygen or (25X32)/2 g= 400 g O₂ is required. So to burn 1.15 g of butane the amount of oxygen required is= g= 4.42 g.