The rule is for every 1 gram of sodium chloride you add .1 degree F to the boiling point up too 100 grams or 212.5 degrees F
2C6H14 + 13O2 ---> 6CO2 +14H2O
M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol
86.17 g C6H14 is 1 mole.
2C6H14 + 13O2 ---> 6CO2 +14H2O
from reaction 2 mol 6 mol
from the problem 1 mol 3 mol
M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Answer : 132.0 g CO2