First, let's write down the balanced chemical equation for the reaction:
2Al(s) + Fe2O3(s) = Al2O3(s) + 2Fe(l)
Next, we need to calculate the heat released in the reaction using the enthalpies of formation of the reactants and products. The enthalpy change for the reaction can be calculated as:
ΔH = ∑(ΔHf(products)) - ∑(ΔHf(reactants))
The enthalpy of formation of Al2O3 is -1675.7 kJ/mol and Fe2O3 is -825.5 kJ/mol
ΔH = [2(-825.5 kJ/mol) + 2(12.4 kJ/mol)] - [2(-1675.7 kJ/mol)]
ΔH = [-1651 kJ/mol + 24.8 kJ/mol] - [-3351.4 kJ/mol]
ΔH = -1626.2 kJ/mol + 3351.4 kJ/mol
ΔH = 1725.2 kJ/mol
Therefore, the heat released in the reaction is 1725.2 kJ/mol of Al reacted with Fe2O3.
To calculate the heat release per gram of Al, we need to first calculate the molar mass of Al. The molar mass of Al is 26.98 g/mol.
Now, we convert the heat release from kJ/mol to kJ/g:
(1725.2 kJ/mol) / (26.98 g/mol) = 63.98 kJ/g
Therefore, the heat release in kilojoules per gram of Al reacted with Fe2O3 is 63.98 kJ/g.
In reaction below
2Al(s) +Fe2O3(s) = Al2O3(s) +2Fe(l)
Calculate the heat release in kilojoules per gram of Al reacted with Fe2O3 . The standard enthalpy of formation for Fe (l) is 12.4kj/mole.
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