$Ba(OH)_2+H_2SO_4\to BaSO_4↓+2H_2O$
$n_{Ba(OH)_2}=0,4.0,2=0,08(mol)$
$n_{H_2SO_4}=0,3.0,25=0,075(mol)$
Xét tỉ lệ: $\dfrac{n_{Ba(OH)_2}}{1}>\dfrac{n_{H_2SO_4}}{1}(0,08>0,075mol)$
$→Ba(OH)_2$ dư $0,08-0,075=0,005(mol)$
Theo PTHH: $n_{BaSO_4}=n_{H_2SO_4}$
$→n_{BaSO_4}=0,075(mol)\\→m_{BaSO_4}=0,075.233=17,475g\\→C$