What mass of carbon dioxide (co2) can be produced from 86.17 grams of c6h14 and excess oxygen?

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