Question

the reaction for the combustion of propane is

mathrm{c}_{3} mathrm{h}_{8}(mathrm{~g})+5 mathrm{o}_{2}(mathrm{~g}) \to 3 mathrm{co}_{2}(mathrm{~g})+4 mathrm{h}_{2} mathrm{o}(mathrm{l}) quad delta h=-526.3 \frac{mathrm{kcal}}{mathrm{mol}}

how much heat is released when ( 56.8 mathrm{~g} ) of propane is burned?

heat: kcal

Explanation:

Step1: Calculate the molar mass of propane ($C_3H_8$)

The molar mass of $C$ is $12.01\ \frac{g}{mol}$ and of $H$ is $1.008\ \frac{g}{mol}$.
For $C_3H_8$: $M=(3\times12.01 + 8\times1.008)\ \frac{g}{mol}=(36.03+8.064)\ \frac{g}{mol}=44.094\ \frac{g}{mol}$

Step2: Calculate the number of moles of propane

Use the formula $n=\frac{m}{M}$, where $m = 56.8\ g$ and $M = 44.094\ \frac{g}{mol}$
$n=\frac{56.8\ g}{44.094\ \frac{g}{mol}}\approx1.29\ mol$

Step3: Calculate the heat released

Given $\Delta H=- 526.3\ \frac{kcal}{mol}$ (heat released per mole of $C_3H_8$)
The heat released $q=n\times|\Delta H|$
$q = 1.29\ mol\times526.3\ \frac{kcal}{mol}\approx679\ kcal$

Answer:

$679$