# 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

Jul 26, 2017

Well, apparently,

${m}_{{C}_{3} {H}_{8}} = \text{24.8 kg}$

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

${q}_{r x n} = \Delta {H}_{r x n}$

I assume your $\Delta {H}_{r x n}$ units are not correct and should be $\text{kJ/mol}$; otherwise, there would not be any point in knowing the mass of the reactant.

Define $\Delta {\overline{H}}_{r x n} = \frac{\Delta {H}_{r x n}}{n} _ \left({C}_{3} {H}_{8}\right)$, where ${n}_{{C}_{3} {H}_{8}}$ is mols of ${\text{C"_3"H}}_{8}$. This means...

${n}_{{C}_{3} {H}_{8}} \Delta {\overline{H}}_{r x n} = \Delta {H}_{r x n} = {q}_{r x n}$

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

$= - 1.25 \times {10}^{6}$ $\text{kJ}$

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")

$=$ $\text{563.09 mols}$

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

$\textcolor{b l u e}{{m}_{{C}_{3} {H}_{8}}} = 563.09 \cancel{{\text{mols C"_3"H"_8) xx ("44.1 g C"_3"H"_8)/cancel("1 mol C"_3"H}}_{8}}$

$=$ ${\text{24832.20 g C"_3"H}}_{8}$

$= \textcolor{b l u e}{{\text{24.8 kg C"_3"H}}_{8}}$

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