Đáp án:

 Al

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2M + 3C{l_2} \to 2MC{l_3}\\
nM = nMC{l_3}\\
\dfrac{{10,8}}{{MM}} = \dfrac{{53,4}}{{MM + 35,5 \times 3}}\\
 \Rightarrow MM = 27g/mol\\
 \Rightarrow M:Al\\
b)\\
\end{array}\)

Kết tủa lớn nhất khi kết tủa không tan và  $AlCl_3$ phản ứng hết

\(\begin{array}{l}
2AlC{l_3} + 3Ca{(OH)_2} \to 3CaC{l_2} + 2Al{(OH)_3}\\
nCa{(OH)_2} = 0,4 \times \dfrac{3}{2} = 0,6\,mol\\
V = \dfrac{{0,6}}{{0,1}} = 6l
\end{array}\)

Không có kết tủa khi kết tủa tan hết

\(\begin{array}{l}
2AlC{l_3} + 3Ca{(OH)_2} \to 3CaC{l_2} + 2Al{(OH)_3}\\
0,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,4\\
2Al{(OH)_3} + Ca{(OH)_2} \to Ca{(Al{O_2})_2} + 4{H_2}O\\
0,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2\\
nCa{(OH)_2} = 0,6 + 0,2 = 0,8\,mol\\
V = \dfrac{{0,8}}{{0,1}} = 8l
\end{array}\)

-7,8g kết tủa

TH1: kết tủa chưa tan

\(\begin{array}{l}
nAl{(OH)_3} = \dfrac{{7,8}}{{78}} = 0,1\,mol\\
2AlC{l_3} + 3Ca{(OH)_2} \to 3CaC{l_2} + 2Al{(OH)_3}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\\
VCa{(OH)_2} = \dfrac{{0,15}}{{0,1}} = 1,5l
\end{array}\)

TH2: kết tủa tan 1 phần

\(\begin{array}{l}
2AlC{l_3} + 3Ca{(OH)_2} \to 3CaC{l_2} + 2Al{(OH)_3}\\
\,0,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
2Al{(OH)_3} + Ca{(OH)_2} \to Ca{(Al{O_2})_2} + 4{H_2}O\\
0,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\\
nAl{(OH)_3}\,\tan  = 0,4 - 0,1 = 0,3\,mol\\
nCa{(OH)_2} = 0,15 + 0,6 = 0,75\,mol\\
V = \dfrac{{0,75}}{{0,1}} = 7,5l
\end{array}\)