Answer : The heat given off to the surroundings will be, -155.38 KJ
Solution : Given,
[tex]\Delta H^o_{rxn}=-849.1KJ/mole[/tex]
Mass of aluminium = 9.873 g
Molar mass of aluminium = 27 g/mole
First we have to calculate the moles of aluminium.
[tex]\text{ Moles of Al}=\frac{\text{ Mass of Al}}{\text{ Molar mass of Al}}=\frac{9.873g}{27g/mole}=0.366moles[/tex]
The given balanced reaction is,
[tex]2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe[/tex]
From the reaction, we conclude that
2 moles of aluminum releases heat energy = -849.1 KJ
0.366 moles of aluminium releases heat energy = [tex]\frac{-849.1KJ}{2moles}\times 0.366moles=-155.38KJ[/tex]
Therefore, the heat given off to the surroundings will be, -155.38 KJ
