\(n_{Ba\left(OH\right)_2}=0,5.0,1=0,05\left(mol\right)\)
\(H_2SO_4+Ba\left(OH\right)_2-->BaSO_4+2H_2O\)
a, theo pt : n Ba(OH)2 = n H2SO4= nBaSO4
\(m_{muoitt}=m_{BaSO_4}=0,05.233=11,65\left(g\right)\)
b, \(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
c, \(2HCl+Ba\left(OH\right)_2-->BaCl_2+2H_2O\)
\(n_{HCl}=2m_{Ba\left(OH\right)_2}=2.0,05=0,1\left(mol\right)\)
\(=>V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,1}{1}=0,1\left(l\right)=100\left(ml\right)\)