how much energy is required to convert 1.00 kg of iron III to atomic iron?
Equation: Fe2O3 + 3CO ---> 2 Fe +3 CO2 delta H is 131kJ
delta H is 131kj/moleFe2O3
how many moles is 1000grams of the compound?
delta H is 21 kj sorry
1000 grams is 6.261740764
moles
so heat generated is 6.26moles*21kj/mole
To determine how much energy is required to convert 1.00 kg of iron III (Fe2O3) to atomic iron (Fe), we can use the given equation and the molar mass of iron.
First, let's calculate the molar mass of Fe2O3:
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Fe2O3 = (2 * 55.85 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol
Next, we need to convert the mass of Fe2O3 from kg to grams:
1.00 kg = 1000 g
Now, let's calculate the number of moles of Fe2O3:
Number of moles = mass / molar mass
Number of moles of Fe2O3 = 1000 g / 159.69 g/mol = 6.2614 moles
According to the balanced equation, 2 moles of Fe2O3 react to form 2 moles of Fe. Therefore, the number of moles of Fe produced is also 6.2614 moles.
Now, we can calculate the amount of energy required to convert the moles of Fe2O3 to Fe by using the given enthalpy change (delta H) value:
Energy = number of moles * delta H
Energy = 6.2614 moles * 131 kJ/mol
Finally, let's calculate the energy required in kilojoules:
Energy = 6.2614 moles * 131 kJ/mol = 819.0814 kJ
Therefore, the energy required to convert 1.00 kg of iron III to atomic iron is approximately 819.0814 kJ.