Answer : -69.286 KJ heat energy is liberated.
Solution : Given,
Standard enthalpy of formation : When one mole of a substance is formed from its constituent elements.
[tex]\Delta H_f=962.3KJ/mole[/tex] (for 1 mole)
Mass of Mn = 7.9 g
Molar mass of Mn = 54.93 g/mole
First we have to calculate the moles of Mn.
[tex]\text{ Moles of Mn}=\frac{\text{ Mass of Mn}}{\text{ Molar mass of Mn}}=\frac{7.9g}{54.93g/mole}=0.144moles[/tex]
The formation reaction of [tex]Mn_2O_3[/tex] is represented as,
[tex]4Mn+3O_2\rightarrow 2Mn_2O_3[/tex]
For this reaction, the standard enthalpy of formation is,
[tex]\Delta H_f=2\times 962.3KJ/mole=1924.6KJ/mole[/tex]
From the reaction, we conclude that
4 moles of Mn releases heat energy = -1924.6 KJ
0.144 moles of Mn releases heat energy = [tex]\frac{-1924.6KJ}{4mole}\times 0.144mole=-69.286KJ[/tex]
Therefore, -69.286 KJ heat energy is liberated.
