\(n_{AgNO_3}=0,15.1=0,15\left(mol\right)\)
\(n_{BaCl_2}=0,35.1=0,35\left(mol\right)\)
\(AgNO_3\rightarrow Ag^++NO_3^-\)
0,15---->0,15--->0,15
\(BaCl_2\rightarrow Ba^{2+}+2Cl^-\)
0,35---->0,35--->0,7
\(Ag^++Cl^-\rightarrow AgCl\)
0,15-->0,15-->0,15
mAgCl = 0,15.143,5 = 21,525 (g)
dd sau pư chứa \(\left\{{}\begin{matrix}Cl^-:0,55\left(mol\right)\\Ba^{2+}:0,35\left(mol\right)\\NO_3^-:0,15\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left[Cl^-\right]=\dfrac{0,55}{0,15+0,35}=1,1M\\\left[Ba^{2+}\right]=\dfrac{0,35}{0,15+0,35}=0,7M\\\left[NO_3^-\right]=\dfrac{0,15}{0,15+0,35}=0,3M\end{matrix}\right.\)