nOH-= 0.4 mol
nCO2= 0.3 mol
tỉ lệ nOH-/nCO2= 4/3 => sinh ra 2 muối
CO2 + OH- --> HCO3-
a..........a
CO2+ 2OH- ---> CO3 2- + H2O
b............2b..........b
ta có :
a+b= 0.3
a+2b= 0.4=> b=0.1
nCa2+ = 0.15 mol
=> nCaCO3= nCO3 2- = 0.1 mol => mkt = 10gam