Ba(NO3)2 +H2SO4 --> BaSO4 + 2HNO3 (1)
gọi nBa(NO3)2=x(mol)
=> nBaSO4=x(mol)
=> mBa(NO3)2=261x(g)
mBaSO4=233x(mol)
Mà m 2 muối khác nhau 12,6g
=> 261x -233x=12,6
=> x=0,45(mol)
=> a=117,45(g)
mBaSO4=c=104,85(g)
nH2SO4=0,45(mol)
=>mH2SO4=44,1(g)
=> mddH2SO4=220,5(g)
mà m ddH2SO4 lấy dư 20%
=> mddH2SO4 THỰC TẾ=\(\dfrac{220,5}{120}.100=183,75\left(g\right)\)