a) \(2AgNO_3+BaCl_2\rightarrow Ba\left(NO_3\right)_2+2AgCl\)
b)
\(n_{AgCl}=\dfrac{5,74}{143,5}=0,04\left(mol\right)\)
Theo PTHH: \(\left\{{}\begin{matrix}n_{AgNO_3}=n_{AgCl}=0,04\left(mol\right)\Rightarrow m_1=0,04.170=6,8\left(g\right)\\n_{BaCl_2}=\dfrac{1}{2}n_{AgCl}=0,02\left(mol\right)\Rightarrow m_2=0,02.208=4,16\left(g\right)\end{matrix}\right.\)