`a)`
\(\left\{{}\begin{matrix}m_{AgNO_3}=\dfrac{34}{100}.100=34\left(g\right)\\m_{BaCl_2}=\dfrac{20,8}{100}.0=10,4\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{AgNO_3}=\dfrac{34}{170}=0,2\left(mol\right)\\n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\end{matrix}\right.\)
PTHH: \(2AgNO_3+BaCl_2\rightarrow2AgCl\downarrow+Ba\left(NO_3\right)_2\)
ban đầu 0,2 0,05
sau pư 0,1 0 0,1 0,05
`b)`
\(m_{dd}=100+50-0,1.143,5=135,65\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}C\%_{AgNO_3\left(dư\right)}=\dfrac{0,1.170}{135,65}.100\%=12,53\%\\C\%_{Ba\left(NO_3\right)_2}=\dfrac{0,05.261}{135,65}.100\%=9,62\%\end{matrix}\right.\)