\(n_{H_2SO_4}=\dfrac{150,19,6\%}{98}=0,3mol\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
0,3 0,3 0,3 0,6 ( mol )
\(m_{BaCl_2}=0,3.208=62,4g\)
\(x\%=\dfrac{62,4}{200}.100=31,2\%\)
\(C\%_{BaSO_4}=\dfrac{0,3.233}{200+150}.100=19,97\%\)
\(C\%_{HCl}=\dfrac{0,6.36,5}{200+150}.100=6,25\%\)