13.2 gallons=49.9 liters
the gas is the octan C8H18
the combustion of Octan equation is
C8H18 + 25/2 O2-------------------> 8CO2 + 9 H2O
1mole 8 moles
n(C8H18)=a n(CO2)=b
so
1 mole/a = 8moles /b
and b= 8molesx a
we must find the value of a, molar volume of a gas is v= 22.4l/mol=V/a, so a=V/v= 49.9/22.4= 2.23 moles
and b=8 x a=8x 2.23=17.85 moles
M(CO2)=m(CO2)/n(CO2)====> m=n x M=17.85 * (12+2x16)= 785.4 g
m(CO2)=785.4g = 1.57 pound