What is the mass of octane burnt and calculating the energy released? Full Question in the description box below.
With question a, I just need someone to explain to me in detail, how to receive the Molar mass and mole of Octane.
Whereas question b, I'm confused on how to calculate the energy released for octane.
With question a, I just need someone to explain to me in detail, how to receive the Molar mass and mole of Octane.
Whereas question b, I'm confused on how to calculate the energy released for octane.
2 Answers
I got:
#a)# #m_(oct) = "1.05 g"#
#b)# #|DeltaH_(rxn)| = 3.33 xx 10^6 "kJ"#
This is really a unit conversion.
You were given:
#2"C"_8"H"_18(l) + 25"O"_2(g) -> 16"CO"_2(g) + 18"H"_2"O"(l)# ,
#DeltaH_(rxn) = -"10900 kJ/mol"#
I'm assuming the "per mol" is "per mol of octane". That would mean that for this reaction as-written, which contains
#DeltaH_(rxn) = -"10900 kJ"/"1 mol octane"#
or
#DeltaH_(rxn) = -"21800 kJ"/"2 mols octane"#
(Some books or online questions might simply have written
If you want to yield
The yielded energy is used for something else, so it's positive.
Since it is really the same reaction at different scales, and something in units of
#"10900 kJ"/("1 mol octane") = ("100 kJ")/(x" mols octane")#
Therefore, the mols of octane involved are:
#100 cancel"kJ" xx ("1 mol octane")/(10900 cancel"kJ")#
#=# #"0.00917 mols octane"#
This gives a needed mass of:
#color(blue)(m_(oct)) = 0.00917 cancel("mols C"_8"H"_18) xx (8 cdot "12.011 g" + 18 cdot "1.0079 g C"_8"H"_18)/(cancel("1 mol C"_8"H"_18))#
#=# #color(blue)("1.05 g octane")# to three significant figures.
Given the density and volume, you can find the mass involved:
#50.0 cancel"L octane" xx (1000 cancel"mL")/cancel"1 L" xx "0.698 g octane"/cancel"mL octane"#
#=# #"34900 g octane"#
From this mass, we can work backwards from problem
#34900 cancel"g octane" xx ("1 mol C"_8"H"_18)/(cancel(8 cdot "12.011 g" + 18 cdot "1.0079 g C"_8"H"_18))#
#=# #"305.52 mols octane"#
So, the heat energy released (made into a positive quantity by the wording) is:
#color(blue)(|DeltaH_(rxn)|) = "10900 kJ"/cancel"1 mol octane" xx 305.52 cancel"mols octane"#
#= color(blue)(3.33 xx 10^6 "kJ")# to three significant figures.
a 2.10 g; b 1.66 GJ.
Explanation:
a. Mass of octane
Note: The value
b. Energy released