\(m_{BaCl_2}=\dfrac{200.5,2}{100}=10,4\left(g\right)\\ \rightarrow n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{58,8.20}{100}=11,76\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{11,76}{98}=0,12\left(mol\right)\)
\(PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
- Ta có: 0,05/1 < 0,12/1
=> BaCl2 hết, H2SO4 dư.
=> Các chất trong dd sau phản ứng là H2SO4 (dư) và HCl.
\(n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \rightarrow m_{BaSO_4}=0,05.233=11,65\left(g\right)\)
Ta có: \(m_{ddsau}=200+58,8-11,65=247,15\left(g\right)\)
\(m_{H_2SO_4\left(dư\right)}=11,76-\left(0,12-0,05\right).98=4,9\left(g\right)\)
\(n_{HCl}=2.0,05=0,1\left(mol\right)\\ \rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)
=> \(C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{4,9}{247,15}.100\approx1,983\%\)
\(C\%ddHCl=\dfrac{3,65}{217,15}.100\approx1,477\%\)