PTHH :
CaCO3 \(\underrightarrow{t^o}\) CaO + CO2
2CO2 + Ba(OH)2 \(\rightarrow\) Ba(HCO3)2
Ta có : nCaCO3 = 30 : 100 = 0,3(mol)
=> nCO2 = 0,3 (mol)
nBa(HCO3)2 = 31,08 : 259 = 0,12(mol)
Ta thấy :
\(\dfrac{0,3}{2}>\dfrac{0,12}{1}\)
=> CO2 dư
=> nBa(OH)2 = 0,12 (mol)
Ta có CT :
CM = \(\dfrac{n}{V}\)= \(\dfrac{0,12}{0,8}=0,15M\)
=> a = 0,15M