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.
the final concentration of NaI solution in 60 grams/litre.
Explanation:
Given that:
Initial concentration of NaI solution M1 = 0.2 M
initial volume of NaI V1 = 2 L
Final volume V2 = 1 Litre
Final molarity=?
concentration in grams/litre = ?
molar mass of NaI = 150 gram/mole
For dilution following formula is used:
M1 V1 = M2V2
putting the values in the equation
0.2 X 2 = 1 X M2
M2 = 0.4
For concentration in grams/litre, formula used
molarity =
mass = 0.4 x 150
= 60 grams
So, 60 grams of NaI will be present in final solution of NaI after evaporation.
The concentration is 60 grams/ L (as volume got reduced to 1 litre from 2 litres)