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How much octane could a #0.500*L# volume of air completely combust?

Answer 1

Caleb Adams

Well we need to find TWO quantities..........and finally get a mass of #11.9*mg# with respect to propane.

#(i)# #"Moles of dioxygen"=(PV)/(RT)=((749.1*mm*Hg)/(760*mm*Hg*atm^-1)xx0.1048*L)/(0.0821*(L*atm)/(K*mol)xx328.7*K)#

(Note that I did modify the dioxygen volume to reflect its proportional volume in air.)

#=3.383xx10^-3*mol# WITH RESPECT TO DIOXYGEN.

In order to achieve stoichiometric equivalency, we then follow the stoichiometric reaction.

#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#

Therefore, the formula for "Mass of propane" is (3.383x10^-3mol)/(25/2)x44.10g*mol^-1.

#=11.9*mg#..........

It should be noted that we made the somewhat implausible assumption that combustion would occur completely.

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Answer 2

Axel Alvarado

To calculate the amount of octane that could completely combust in a 0.500 L volume of air, we need to know the stoichiometric ratio of air to octane in the combustion reaction, which is approximately 14.7 parts of air to 1 part of octane by mass. From there, we can determine the mass of octane that would be required to combust completely in the given volume of air.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.