a) PTHH: \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
Ta có: \(n_{BaO}=\frac{7,65}{153}=0,05\left(mol\right)=n_{Ba\left(OH\right)_2}\) \(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\frac{0,05}{0,5}=0,1\left(M\right)\)
b) PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)
Theo PTHH: \(n_{Ba\left(OH\right)_2}=n_{BaSO_4}=0,05mol\)
\(\Rightarrow m_{BaSO_4}=0,05\cdot233=11,65\left(g\right)\)