BaCO₃=>BaO+CO₂

n(BaCO₃)=19,7/197=0,1mol

a)

BaO+H₂O=>Ba(OH)₂

m(Ba(OH)₂)=0,1×171=17,1g

mdd=120+15,3=135,3g

C%=17,1/135,3×100%=12,64%

CM=0,1/0,1353=0,739M

b)

n(NaOH)=100×5%/40=0,125mol

CO₂+NaOH=>NaHCO₃

n(NaOH dư)=0,125-0,1=0,025mol

m(NaOH dư)=1g

m(NaHCO₃)=0,1×84=8,4g

mdd=100+4,4=104,4g

C%(NaHCO₃)=8,05%

C%(NaOH)=0,96%