The complete combustion of octane, a component of gasoline, is represented by the equation: 2 c8h18(l) + 25 o2(g) →16 co2(g) + 18 h2o(l) how many liters of co2(g), measured at 63.1°c and 688 mmhg, are produced for every gallon of octane burned? (1 gal = 3.785 l; density of c8h18(l) = 0.703 g/ml)

5670 liter

Explanation:

1) Chemical equation (given):

2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(l)

2) Mole ratio:

2 mol C₈H₁₈(l) : 16 mol CO₂(g)

3) C₈H₁₈ (l) moles:

Molar mass: 114.2285 g/mol (taken from a table or internet)

Volume C₈H₁₈ = 1 galon = 3.785 liter (given)density = mass / volume ⇒ mass = density × volumemass = 0.703 g/ ml × 3785 ml  = 2,661 gmoles = mass in grams / molar mass = 2,661 g / 114.2285 g/mol = 23.3 mol

4) Proportion:

2 mol C₈H₁₈(l) / 16 mol CO₂(g) = 23.3 mol C₈H₁₈(l) / xx = 186 mol CO₂ (g)

5) Ideal gas equation:

pV = nRT

Substitute with:

n = 186 molR = 0.08206 atm-liter / mol-KT = 63.1 + 273.15 K = 336.25 Kp = 688 mmHg × 1 atm/760 mmHg = 0.905 atm

Solve for V:

V = 186 mol × 0.08206 atm-liter / K-mol × 336.25K / 0.905 atmV = 5671 liter = 5670 liter (using 3 significant figures) ← answer

Ido not understand. 

While unproven, it uses facts to support part of a conclusion. it is still factual and has predictive power, however.