\(m_{BaCl_2}=\dfrac{200\cdot15,6\%}{100\%}=31,2g\Rightarrow n_{BaCl_2}=0,15mol\)
\(n_{H_2SO_4}=\dfrac{150}{98}=\dfrac{75}{49}mol\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,15 \(\dfrac{75}{49}\) 0,15 0,3
a)\(m_{\downarrow}=0,15\cdot233=34,95g\)
b)\(m_{HCl}=0,3\cdot18=5,4g\)
\(m_{ddBaSO_4}=200+150-5,4=344,6g\)
\(C\%=\dfrac{5,4}{344,6}\cdot100\%=1,57\%\)