\(BaCl_2+H_2SO_4\xrightarrow[]{}BaSO_4+2HCl\)
(mol) 0,05.............0,05..............0,05..........0,1
\(n_{BaCl_2}=\dfrac{104.10}{100.208}=0,05\left(mol\right)\)
\(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)
Theo BTKL: \(m_{ddH_2SO_4}+m_{ddBaCl_2}=m_{ddsau}+m_{BaSO_4}\)
⇒ \(m_{ddsau}=\dfrac{4,9.100}{7}+104-0,05.233=162,35\left(g\right)\)
\(C\%_{HCl}=\dfrac{0,1.36,5}{162,35}.100=2,25\%\)