Question

Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 6.9 g of octane is mixed with 44.9 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant figures.

Answer 1

octane combustion equation is:
2 C8H18 + 25 O2 ---> 16 CO2 +18 H2O

moles of C8H18 = mass / molar mass
= 6.9 / (8*12 + 18*1)
= 0.0605 mol

moles of O2 = mass/molar mass
= 44.9/32
=1.403 mol

from reaction given,
2 mol of C8H18 needs 25 mol of O2
so,
0.0605 mol of C8H18 will need = (25/2)*0.0605 mol O2
=0.756 mol of O2
so,
C8H18 is limiting reagent
2 mol of C8H18 forms 16 mol of CO2
moles of CO2 formed = (16/2)*0.0605
=0.484 mol

molar mass of CO2 = 44 g/mol
mass of CO2 = number of moles * molar mass
= 0.484 * 44
= 21.3 g
Answer:21 g