When a sample of ca(s) loses 1 mole of electrons in a reaction with a sample of o2(g) the oxygen?

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.

It is 4.92 e-4 sorry if i am wrong