Question

Calculate the enthalpy change for the reduction of iron (III) oxide using carbon in a thermite reaction if 12.5 grams of iron (III) oxide reacts with excess carbon, and the resulting iron has a mass of 7.5 grams. Assume all heat evolved is absorbed by the iron and that the reaction occurs at constant pressure.

Answer 1

To calculate the enthalpy change for the reduction of iron  III  oxide using carbon in a thermite reaction, we need to follow these steps:1. Write the balanced chemical equation for the reaction.2. Calculate the moles of iron  III  oxide and iron produced.3. Determine the enthalpy change for the reaction using stoichiometry and the enthalpy of formation values.Step 1: Balanced chemical equationFe2O3  s  + 3C  s   2Fe  s  + 3CO  g Step 2: Calculate the moles of iron  III  oxide and iron producedMolar mass of Fe2O3 =  2  55.85  +  3  16.00  = 111.7 + 48 = 159.7 g/molMoles of Fe2O3 = mass / molar mass = 12.5 g / 159.7 g/mol = 0.0783 molFrom the balanced equation, 1 mol of Fe2O3 produces 2 mol of Fe.Moles of Fe produced = 0.0783 mol  2 = 0.1566 molStep 3: Determine the enthalpy change for the reactionWe need the standard enthalpy of formation values for the reactants and products. The values are as follows:Hf  Fe2O3  = -824.2 kJ/molHf  C  = 0 kJ/mol  since it is in its standard state Hf  Fe  = 0 kJ/mol  since it is in its standard state Hf  CO  = -110.5 kJ/molUsing the formula H =  Hf  products  -  Hf  reactants , we get:H = [ 2  0  +  3  -110.5 ] - [ 1  -824.2  +  3  0 ]H = [-331.5] - [-824.2]H = 492.7 kJ/molNow, we need to find the enthalpy change for the 0.0783 mol of Fe2O3 reacted:H = 0.0783 mol  492.7 kJ/mol = 38.56 kJThe enthalpy change for the reduction of 12.5 grams of iron  III  oxide using carbon in a thermite reaction is 38.56 kJ.