Question

Using the balanced equation shown below, what is the mass of C3H8 that must react in order to release 1.25×10^6 kJ of heat? ΔHrxn = –2219.9 kJ

Answer 1

Well, apparently,

`m_(C_3H_8) = "24.8 kg"`

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Reactions tend to occur at constant pressure, so consequently, we write that

`q_(rxn) = DeltaH_(rxn)`

I assume your `DeltaH_(rxn)` units are not correct and should be `"kJ/mol"`; otherwise, there would not be any point in knowing the mass of the reactant.

Define `DeltabarH_(rxn) = (DeltaH_(rxn))/n_(C_3H_8)`, where `n_(C_3H_8)` is mols of `"C"_3"H"_8`. This means...

`n_(C_3H_8)DeltabarH_(rxn) = DeltaH_(rxn) = q_(rxn)`

`= n_(C_3H_8) xx (-"2219.9 kJ"/("mol C"_3"H"_8))`

`= -1.25 xx 10^6` `"kJ"`

Therefore, this many mols of propane reacted:

`n_(C_3H_8) = -1.25 xx 10^6 cancel"kJ" xx ("1 mol C"_3"H"_8)/(-2219.9 cancel"kJ")`

`=` `"563.09 mols"`

...wow, that's huge... Well, in that case...

`color(blue)(m_(C_3H_8)) = 563.09 cancel("mols C"_3"H"_8) xx ("44.1 g C"_3"H"_8)/cancel("1 mol C"_3"H"_8)`

`=` `"24832.20 g C"_3"H"_8`

`= color(blue)("24.8 kg C"_3"H"_8)`

I would not want to be at this factory... they're combusting kilos of propane!