**Calculating Energy Transferred**

The reaction of iron with oxygen from the air provides the energy transferred by the hot pack described in Conceptual Exercise 6.12. Assuming that the iron is converted to iron(III) oxide, how much heating can be provided by a hot pack that contains a tenth of a pound of iron? The thermochemical expression is

2 Fe( s ) + \frac{3}{2}O_2(g) → Fe_2O_3(s) ΔH° = – 824.2 kJ

Step-by-Step

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– 335 kJ

**Strategy and Explanation** Begin by calculating how many moles of iron are present. A pound is 454 g, so

Amount of iron = 0.100 lb × \frac{454 g}{1 lb} × \frac{1 mol Fe}{55.84 g} = 0.8130 mol Fe

Then use a thermostoichiometric factor to calculate the energy transferred. The appropriate factor is 824.2 kJ transferred to the surroundings per 2 mol Fe, so

Energy transferred = 0.8130 mol Fe × \frac{- 824.2 kJ}{2 mol Fe} = – 335 kJ

Thus, 335 kJ is transferred by the reaction to heat your fingers.

**Reasonable Answer Check** A tenth of a pound is about 45 g, which is a bit less than the molar mass of iron, so we are oxidizing less than a mole of iron. Two moles of iron gives about 800 kJ, so less than a mole should give less than 400 kJ, which makes 335 kJ a reasonable value. The sign should be negative because the enthalpy of the hand warmer (system) should go down when it transfers energy to your hand.

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