Question

when 1.25 gram of iron reacts with sulphur to form Fe(II) sulphide 1.24KJ of heat is released. Calculate the enthalpy of the reaction in KJ per mole.

Answer 1

$Fe+ S \Rightarrow FeS$

As the energy given is for the 1 mole of iron.

For the enthalpy of the reaction we have to find the moles of iron in 1.25 gm and then multiply it with the given energy.

Now moles of Fe in 1.25 gm of iron $=$ $\dfrac{1.25}{55.845}=0.022 moles$

So the enthaply of the reaction is $=0.022 \times1.24 kJ=0.027 kJ$