Đáp án:
CuCl2
Giải thích các bước giải:
\(\begin{array}{l}
R{X_2} + 2AgN{O_3} \to 2AgX + R{(N{O_3})_2}\\
Fe + R{X_2} \to R + FeC{l_2}\\
nAgX = \frac{{5,74}}{{108 + MX}}\,\,mol\\
= > nR{X_2} = \frac{{2,87}}{{108 + MX}}mol\\
nFe = nR = \frac{{2,87}}{{108 + MX}}\,mol\\
MR \times \frac{{2,87}}{{108 + MX}} - 56 \times \frac{{2,87}}{{108 + MX}} = 0,16\\
MR \times 2,87 - 160,72 = 17,28 + 0,16MX\\
17,9375MR - MX = 1112,5\\
= > MX = 35,5MR = 64\\
= > X:Cl,R:Cu\\
= > CTHH:CuC{l_2}
\end{array}\)