The balanced combustion reaction of octane ( which is a component of gasoline) can be written as:
C₈H₈ + [tex]\frac{25}{2}[/tex] O₂→ 8 CO₂ + 9 H₂O.
(a) For combustion of one mole of butane, [tex]\frac{25}{2}[/tex] moles of oxygen is required. So, for combustion of 1.60 mole of butane, Oxygen required is [tex]\frac{25 X 1.60}{2}[/tex] 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= [tex]\frac{400 X 1.15}{104}[/tex] g= 4.42 g.
(a) number of moles of O2 needed to 1.60 C8H18 is
=mole ratio of C8H18:O2=2:25
mole of O2 is therefore=1.6 x25/2=20 moles
(B) number of moles of CO2 produced
mole ratio of C8H18:CO2=2:16
moes of CO2 is therefore=0.19 x16/2=1.52 moes
(c) the grams of O2
moles of C8H18=1.15g/114 g/mol=0.01 moles
mole of C8H18:O2=2:25
moles of O2 is therefore=0.01 x25/2=0.125moles
mass=0.125moles x32 g/mol=4grams
