Question

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.

Answer 1

I got:

`a)` `m_(oct) = "1.05 g"`

`b)` `|DeltaH_(rxn)| = 3.33 xx 10^6 "kJ"`

---

`a)`

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 `"2 mols"` of octane, the enthalpy involved can also be written as:

`DeltaH_(rxn) = -"10900 kJ"/"1 mol octane"`

or

`DeltaH_(rxn) = -"21800 kJ"/"2 mols octane"`

(Some books or online questions might simply have written `-"21800 kJ"` instead of this, so watch for that.)

If you want to yield `"100 kJ"` of energy, since energy scales with the size of the system, or in this case how much mass of octane is burned, you should expect much less than `"2 mols"` of octane is burned.

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 `"kJ/mol"` is intensive, these two values in `"kJ/mol"` must be equal:

`"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.

`b)`

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 `(a)`, basically. Given the `DeltaH` value, we can again use that as a conversion factor.

`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.

Answer 2

a 2.10 g; b 1.66 GJ.

Explanation:

a. Mass of octane

Note: The value `ΔH = "-10 900 kJ·mol"^"-1"` is for 1 mol of reaction, i.e. for 2 mol of octane.

`M_text(r):color(white)(m)114.23`
`color(white)(mmm)"2C"_8"H"_18 + "25O"_2 → "16CO"_2 + "18H"_2"O" + "10 900 kJ"`

`m_text(octane) = 100 color(red)(cancel(color(black)("kJ"))) × "228.46 g"/("10 900" color(red)(cancel(color(black)("kJ")))) = "2.10 g"`

b. Energy released

`m_text(octane) = 50.0 color(red)(cancel(color(black)("L"))) × (1000 color(red)(cancel(color(black)("mL"))))/(1 color(red)(cancel(color(black)("L")))) × "0.698 g"/(1 color(red)(cancel(color(black)("mL")))) = "34 900 g"`

`"Energy released" = "34 900" color(red)(cancel(color(black)("g")))× "10 900 kJ"/(228.46 color(red)(cancel(color(black)("g")))) = 1.66 × 10^6color(white)(l) "kJ" = "1.66 GJ"`