Đáp án:
\(\begin{array}{l}
1.D\\
2.A\\
3.A\\
4.C\\
5.C\\
6.D
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
{n_{Al}} = 0,2mol\\
\to {n_{AlC{l_3}}} = {n_{Al}} = 0,2mol\\
\to {m_{AlC{l_3}}} = 26,7g
\end{array}\)
\(\begin{array}{l}
2.\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{FeC{l_3}}} = 0,04mol\\
\to {n_{Fe}} = {n_{FeC{l_3}}} = 0,04mol\\
\to {m_{Fe}} = 2,24g
\end{array}\)
\(\begin{array}{l}
3.\\
Zn + C{l_2} \to ZnC{l_2}\\
2Al + 3C{l_2} \to 2AlC{l_3}
\end{array}\)
Gọi a và b lần lượt là số mol của Zn và Al
\(\begin{array}{l}
\left\{ \begin{array}{l}
65a + 27b = 11,9\\
136a + 133,5b = 40,3
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,1\\
b = 0,2
\end{array} \right.\\
\to {n_{Zn}} = 0,1mol\\
\to {n_{Al}} = 0,2mol\\
\to {n_{C{l_2}}} = {n_{Zn}} + \dfrac{3}{2}{n_{Al}} = 0,4mol\\
\to {V_{C{l_2}}} = 8,96l
\end{array}\)
\(\begin{array}{l}
4.\\
Na + \frac{1}{2}C{l_2} \to NaCl\\
{n_{NaCl}} = 0,16mol\\
\to {n_{C{l_2}}} = \dfrac{1}{2}{n_{NaCl}} = 0,08mol\\
\to {V_{C{l_2}}} = 1,792l\\
\to {n_{Na}} = {n_{NaCl}} = 0,16mol\\
\to {m_{Na}} = 3,68g\\
\to {m_{Na(tt)}} = \dfrac{{3,68}}{{80}} \times 100 = 4,6g
\end{array}\)
\(\begin{array}{l}
5.\\
2M + xC{l_2} \to 2MC{l_x}\\
{n_M} = {n_{muối}}\\
\to \dfrac{{3,36}}{M} = \dfrac{{9,75}}{{M + 35,5x}} \to M = \dfrac{{56}}{3}x\\
\to \left\{ \begin{array}{l}
x = 1\\
x = 2\\
x = 3
\end{array} \right. \to \left\{ \begin{array}{l}
M = \dfrac{{56}}{3}(loai)\\
M = \dfrac{{112}}{3}(loai)\\
M = 56(Fe)
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
6.\\
2NaOH + C{l_2} \to NaCl + NaClO + {H_2}O\\
{n_{C{l_2}}} = 0,05mol\\
\to {n_{NaOH}} = 2{n_{C{l_2}}} = 0,1mol\\
\to {V_{NaOH}} = 0,1l
\end{array}\)
