chem question please help asap

Question

The combustion of octane, C8H18,proceeds according to the reaction shown.

2 C8H18(1) + 25 O2(g) ------ > 16 CO2(g) + 18 H20(1)

If 5.30 X 10^2 mol of octane combusts, what volume of carbon dioxide is produced at 19 degrees Celsius and .995 atm

Answer 1

Hello, Rokaya,

2 C8H18(1) + 25 O2(g) ------ > 16 CO2(g) + 18 H20(1)

We can see from the balanced equation that 16 moles of CO₂ are produced for every 2 moles of C₈H₁₈. That's a molar ratio of 8 (moles CO₂/mole C₈H₁₈). Since we start with 5.3x10² moles of octane, we get 8 times that amount of CO₂. That comes to 4.024x10³ moles of CO₂.

Use that number in the ideal gas law to caculate volume of CO₂ under the conditions specified. PV=nRT, where T is in Kelvin. Since pressure and volume are given in atm and liters, I'll choose a gas constant, R, that has compatible units. I picked R=0.082058 L*atm/(K*mole).

Use the moles from the first excercise and solve for V, the volume. I get 9.70x10⁴ liters.

Bob