Answer to Question #67381 in General Chemistry for Jordan

Question

402 mol of octane combusts, what volume of carbon dioxide is produced at 40.0 °C and 0.995 atm?

Answer 1

The chemical equation for given reaction is

C8H18 + 12.5O2 → 9H2O + 8CO2

Where 402 mol of octane is
m (C8H18) = n·M(C8H18) = 402·114 = 45828 (g) – mass of octane.
Using chemical reaction, we can determine the mass of carbon dioxide:
45828 g X g
C8H18 + 12.5O2 → 9H2O + 8CO2
114 8·44 = 352
Where 114 g/mol is molar mass of C8H18; 44 g/mol is molar mass of CO2.
Make a proportion:
45828/114 = X/352
So:
X = (45828∙352)/114 = 141504 (g) = m (CO2)
According to Mendeleev – Klapeyron’s equation, we can calculate the volume of CO2 gas:
pV = (m(CO2))/(M(CO2)) ·RT
where p is pressure (p = 0.995 atm); R is gas constant (R = 0,0821 L·atm/mole·K)
Thus, volume of carbon dioxide equals:
V(CO2) = (m(CO2))/(M(CO2)·p) ·RT = 141504/(44·0.995) ·0.0821·(273 + 40) = 83057.8 (L)

Answer: V(CO2) = 83057.8 (L).