Question

The combustion of octane proceeds according to the following reaction.                 2C8H18(l)+25O2(g)→16CO2(g)+18H2O(l)If 474 moles of octane combust, what volume of carbon dioxide is produced at 19.0 degrees Celsius and 0.995 atm?

Answer 1

According to the reaction:

n(CO₂) = $\frac{16}{2}$ n(C₈H₁₈) $= 8 \times 474 = 3792 \;mol$

Ideal Gas Law:

$pV=nRT \\ V= \frac{nRT}{p} \\ p=0.995 \;atm \\ T= 19.0 + 273.15 = 292.15 \;K$

R = 0.08206 L×atm/mol×K

$V= \frac{3792 \times 0.08206 \times 292.15}{0.995} = 91365.58 \;L$

Answer: 91365.58 L