Question

a chemist measures the energy change, delta h, during the reaction: c3h8(g) + 5 o2(g) --> 3co2(g) + 4 h2o(l) delta h=-2220 kj. when 20.000g of propane and 20.000 g of oxygen are combusted, calculate how much heat in kj will be released. attach your work. upload choose a file

Explanation:

Step1: Calculate the molar mass of propane ($C_3H_8$) and oxygen ($O_2$)

The molar mass of $C_3H_8$: $M_{C_3H_8}=(3\times12)+(8\times1)=44\ g/mol$
The molar mass of $O_2$: $M_{O_2}=32\ g/mol$

Step2: Determine the limiting reactant

Moles of $C_3H_8$: $n_{C_3H_8}=\frac{20.000\ g}{44\ g/mol}\approx0.4545\ mol$
Moles of $O_2$: $n_{O_2}=\frac{20.000\ g}{32\ g/mol}=0.625\ mol$
From the balanced equation $C_3H_8(g)+5O_2(g)\to3CO_2(g)+4H_2O(l)$, the mole ratio of $C_3H_8$ to $O_2$ is $1:5$.
For $0.4545\ mol$ of $C_3H_8$, the moles of $O_2$ required is $0.4545\times5 = 2.2725\ mol$. Since we have only $0.625\ mol$ of $O_2$, $O_2$ is the limiting reactant.

Step3: Calculate the heat released

From the balanced equation, $5\ mol$ of $O_2$ releases $2220\ kJ$ of heat.
For $0.625\ mol$ of $O_2$, the heat released $q=\frac{2220\ kJ}{5\ mol}\times0.625\ mol = 277.5\ kJ$

Answer:

$277.5\ kJ$