Respuesta :
Answer : The heat of combustion for cyclopropane in kcal/gram is, 5.95 kcal/g
Explanation : Given,
Heat of reaction for the combustion of cyclopropane = 499.8 kcal/mol
Molar mass of cyclopropane = 42 g/mol
The balanced chemical combustion reaction will be:
[tex]2C_3H_6+9O_2\rightarrow 6CO_2+6H_2O[/tex]
From the balanced chemical reaction we conclude that,
As, the heat of combustion of cyclopropane for 2 moles of cyclopropane = 499.8 kcal
And, the heat of combustion of cyclopropane for 1 mole of cyclopropane = [tex]\frac{1mol}{2mole}\times 499.8kcal=249.9kcal[/tex]
We know that:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
To calculate the heat of combustion in per gram, we divide the given heat of combustion in kcal/mol by the molar mass of cyclopropane, we get:
[tex]\text{Heat of combustion of cyclopropane}=\frac{249.9kcal/mol}{42g/mol}\\\\\text{Heat of combustion of cyclopropane}=5.95kcal/g[/tex]
Hence, the heat of combustion for cyclopropane in kcal/gram is, 5.95 kcal/g
