\(n_{Ba\left(OH\right)_2}=0,2\times0,4=0,08\left(mol\right)\)
\(n_{H_2SO_4}=0,25\times0,3=0,075\left(mol\right)\)
Ba(OH)2 + H2SO4 → BaSO4↓ + 2H2O
Theo pt: \(n_{Ba\left(OH\right)_2}=n_{H_2SO_4}\)
Theo bài: \(n_{Ba\left(OH\right)_2}=\frac{16}{15}n_{H_2SO_4}\)
Vì \(\frac{16}{15}< 1\) ⇒ Ba(OH)2 dư
Theo pT: \(n_{BaSO_4}=n_{H_2SO_4}=0,075\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,075\times233=17,475\left(g\right)\)