Question

If you need 1.00x10^3 kcal of heat, how many grams of Al and Fe2O3 are needed? (8 Al, 1 Fe2O3)

Answer 1

Consider the reaction between Fe₂O₃ and Al

Standard enthalpy of formation of Fe₂O₃==-824.2 kJ/mol

Standard enthalpy of formation of Al₂O₃==-1669.8 kJ/mol

Standard enthalpy of formation of Fe and Al===0 kJ/mol (As they are in their standard state under standard conditions)

Standard enthalpy change for given reaction=

=-1669.8 kJ/mol+0-(-824.2 kJ/mol)-0=-845.6 kJ/mol=-845.6 kJ/mol x 0.239 kCal/kJ=202.10 kCal/mol (1 kJ=0.239 kCal)

So to release 202.10 kCal 1 mol Fe₂O₃ is used up

Molar mass of Fe₂O₃=2 x Molar mass of Fe+3 x Molar mass of O=2 x 55.8 g/mol + 3 x 16 g/mol=111.6 g/mol + 48 g/mol=159.6 g/mol

To release 202.10 kCal heat, 159.6 g Fe₂O₃ is used up

To release 1 kCal heat, 159.6 g/202.10 g Fe₂O₃ is used up

To release 1.00 x 10³ kCal heat, (159.6 g/202.10) x 1.00 x 10³=789.71 g Fe₂O₃ is used up

Again 1 mol Fe₂O₃ reacts with 2 mol Al

i.e. 159.6 g Fe₂O₃ reacts with 2x27 g=54 g Al

1 g Fe₂O₃ reacts with 54 g/159.6 g Al

So 789.71 g Fe₂O₃ reacts with (54 g/159.6) x 789.7 g= 267.20 g Al

So to release 1.00x10³ kCal heat, 789.71 g Fe₂O₃ and 267.20 g Al are required.