# DeltaH_"combustion"^@ "propane"=2220*kJ*mol^-1. What mass of propane is required to yield 76000*kJ of energy?

Mar 14, 2017

$\Delta H {\text{_"rxn}}^{\circ}$ is ALWAYS written per mole of reaction as written. We need approx. $1.5 \cdot k g$ of propane.........

#### Explanation:

And thus the complete combustion of 1 mole of propane, i.e. $44.1 \cdot g$, yields $2200 \cdot k J$ as shown:

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right) + \Delta$

$\text{OR..............}$

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right) + 2200 \cdot k J$

We put the energy output on the PRODUCT side inasmuch the combustion equation is EXOTHERMIC (the negative sign of $\Delta H$ assures us that this is the case). And thus, in effect, we use $\Delta$ as a virtual product in the reaction, and we solve the following quotient:

$\frac{76000 \cdot k J}{2200 \cdot k J \cdot m o {l}^{-} 1} = 34.6 \cdot m o l$, by which we mean $34.6 \cdot m o l$ of reaction as written.

And thus we require $34.6 \cdot m o l$ propane, i.e. $34.6 \cdot m o l \times 44.1 \cdot g \cdot m o {l}^{-} 1 = 1523.5 \cdot g$