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Question

Gasoline (octane) combusts with oxygen gas according to the following reaciton:

2𝐶8𝐻18+25𝑂2⟶16𝐶𝑂2+18𝐻2𝑂

Calculate the volume, in Liters of CO2 (g)  produced when 648.17 g of C8H18 react at 129.75  and 1.83 atm.  (round to hundredths place)   

Answer 1

Hello, Emma,

We'll first calculate the number of moles of CO₂ that will be produced from this combustion reaction. We need to assume that we have sufficient oxygen, and that all of the C₈H₁₈ will react. The balanced equation says we'll obtain 16 moles of carbon dioxide for every 2 moles of C₈H_18, octane. That's a molar ratio of 8 to 1. We need to convert 648.17 grams of octane into moles octane. Divide the mass by the molar mass of C₈H₁₈ to find 5.681 moles of octane. Then we multiply that by 8 to determine the number of moles of carbon dioxide that will be produced. That gives us 45.45 moles CO₂.

Since this is a gas, and the conditions are not STP, we need to use the gas laws to determine the volume of 45.45 moles of a gas at 129.75 (I will assume this is C??) and at 1.83 atm.

Use the ideal gas law, PV = nRT, where n is the moles of gas, R is the gas constant. Temperature must be in Kelvin, so add 273.15 to 129.75C. I use R= 0.08206 L*atm*K^-1mol^-1.

Reaarange the equation to isolate V, the unknown:\:

V = nRT/P

Enter the data, cross out units, where possible, and solve. I get 821.10 liters.

Bob