\(BaCl_2+Na_2SO_4-->BaSO_4+2NaCl\)
\(m_{BaCl_2}=\dfrac{m_{dd}.C\%}{100}=\dfrac{450.20}{100}90g\)
\(n_{BaCl_2}=\dfrac{90}{208}=0.43mol\)
--> \(m_{BaSO_4}=0,43.233=100,19g\)
c, \(m_{ddNaCl}=450+300-100,19=649,81g\)
\(m_{NaCl}=n.M=0,86.58,5=50,31\)
-->\(C\%=\dfrac{50,31}{649,81}.100\%=7,74\%\)