The reaction involved in present case is:
Net Reaction: Ca + 1/2 O2 → CaO. ..................(1)
In terms of oxidation and reduction, the reaction can be shown at
Oxidation: Ca → Ca2+ + 2e- .................(2)
Reduction: 1/2O2 + 2e- → O2-...................(3)
From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.
From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.
Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.
