\(n_{H_2SO_4}=0,6\left(mol\right)\)
\(m_{BaCl_2}=80\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{5}{13}\left(mol\right)\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
Dễ thấy \(H_2SO_4\) dư
\(\Rightarrow n_{BaSO_4\downarrow}=n_{BaCl_2}=\dfrac{5}{13}\left(mol\right)\)
Mà hiệu suất chỉ đạt 80%
\(\Rightarrow n_{BaSO_4\downarrow}=\dfrac{5}{13}.\dfrac{80}{100}=\dfrac{4}{13}\left(mol\right)\)
\(\Rightarrow m_{BaSO_4\downarrow}=71,69\left(g\right)\)