CaCO3 (s) ----> CaO (s) + CO2 (g)
from the above equation :
calculated the amount of CO2 = 44 gms
as CO2 = (1 x 12) + (2 x 16)
= 12 + 36
= 44 gms.
Gram molecular mass of CaCO3 = 40 + 12 + 48
= 100 gms.
hence 25 gm of CaCO3 gives = 44 x 25/100
= 11gms of CO2 liberated.