Đáp án:
\(\begin{array}{l}
1,B\\
2,B\\
3,C\\
4,D\\
5,C\\
6,D\\
7,A\\
8,D\\
9,C\\
10,D\\
11,B\\
12,A\\
13,A\\
14,B\\
15,C\\
16,D\\
17,B\\
18,D\\
19,C\\
20,A\\
21,A\\
22,C\\
23,D
\end{array}\)
Giải thích các bước giải:
7,
\(\begin{array}{l}
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{C{l_2}}} = 0,3mol\\
\to {n_{FeC{l_3}}} = \dfrac{2}{3}{n_{C{l_2}}} = 0,2mol\\
\to {m_{FeC{l_3}}} = 32,5g
\end{array}\)
8,
\(\begin{array}{l}
2A + 3C{l_2} \to 2AC{l_3}\\
{n_A} = {n_{AC{l_3}}}\\
\to \dfrac{{5,4}}{A} = \dfrac{{26,7}}{{A + 35,5 \times 3}}\\
\to A = 27\\
\to Al
\end{array}\)
23,
\(\begin{array}{l}
A + C{l_2} \to AC{l_2}\\
{n_A} = {n_{AC{l_2}}}\\
\to \dfrac{{4,8}}{A} = \dfrac{{19}}{{A + 35,5 \times 2}}\\
\to A = 24\\
\to Mg
\end{array}\)
