0.02L * 0.300 M K2CO2 = 6x10-3 mols K2CO3 = 2 * (6x10-3 mols K(aq)) + 6x10-3 mols CO3 (aq)
0.06L * 0.600 M Ba(NO3)2 = 0.036 mols Ba(NO3)2 = 0.036 mols Ba (aq) + 2* ( 0.036 mols (NO3)2 )
After the reaction, how many mols of Ba are left? In what total volume is it in?