a)
\(2AgNO_3+BaCl_2\rightarrow Ba\left(NO_3\right)_2+2AgCl\downarrow\)
b)
\(n_{AgNO_3}=\dfrac{100.34\%}{170}=0,2\left(mol\right)\)
\(n_{BaCl_2}=\dfrac{50.20,8\%}{208}=0,05\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,05}{1}\) => AgNO3 dư, BaCl2 hết
PTHH: 2AgNO3 + BaCl2 --> Ba(NO3)2 + 2AgCl
0,1<-----0,05-------->0,05----->0,1
mdd sau pư = 100 + 50 - 0,1.143,5 = 135,65 (g)
\(\left\{{}\begin{matrix}m_{Ba\left(NO_3\right)_2}=0,05.261=13,05\left(g\right)\\m_{AgNO_3\left(dư\right)}=\left(0,2-0,1\right).170=17\left(g\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}C\%_{Ba\left(NO_3\right)_2}=\dfrac{13,05}{135,65}.100\%=9,62\%\\C\%_{AgNO_3\left(dư\right)}=\dfrac{17}{135,65}.100\%=12,53\%\end{matrix}\right.\)
