\(m_{CuSO_4}=\dfrac{200\cdot32}{100}=64\left(g\right)\)\(\Rightarrow n_{CuSO_4}=\dfrac{64}{160}=0,4mol\)
\(m_{BaCl_2}=\dfrac{200\cdot10,4}{100}=20,8\left(g\right)\)\(\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1mol\)
\(CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow+CuCl_2\)
0,4 0,1 0,1 0,1
b)\(m_{BaSO_4}=0,1\cdot233=23,3\left(g\right)\)
c)\(m_{CuCl_2}=0,1\cdot135=13,5\left(g\right)\)
\(\Rightarrow m_{ddsau}=200+200-13,5=386,5\left(g\right)\)
\(\Rightarrow C\%=\dfrac{23,3}{386,5}\cdot100\%=6,028\%\)