$\rm a)$

$\rm Mg+2HCl\to MgCl_2+H_2$

$\rm 2Al+6HCl\to 2AlCl_3+3H_2$

$\rm b)$

$\rm n_{HCl}=0,8.2=1,6(mol)$

$\rm n_{H_2}=\dfrac{16,1135}{24,79}=0,65(mol)$

$\rm n_{HCl}>2n_{H_2}$`=>`$\rm HCl\ dư$

$\rm Gọi\ n_{Mg}:a(mol)\ , n_{Al}:b(mol)$

`=>`$\rm 24a+27b=12,9\ (1)$

$\rm n_{H_2}=a+1,5b=0,65\ (2)$

`(1)(2)=>`$\rm a=0,2(mol)\ , b=0,3(mol)$

$\rm m_{Mg}=0,2.24=4,8(g)$

$\rm m_{Al}=12,9-4,8=8,1(g)$

$\rm c)$

$\xrightarrow{\text{BTNT Mg :}}$ $\rm n_{MgCl_2}=n_{Mg}=0,2(mol)$

$\xrightarrow{\text{BTNT Al :}}$ $\rm n_{AlCl_3}=n_{Al}=0,3(mol)$

$\rm C_{M\ MgCl_2}=\dfrac{0,2}{0,8}=0,25M$

$\rm C_{M\ AlCl_3}=\dfrac{0,3}{0,8}=0,375M$

$\rm C_{M\ HCl_{(dư)}}=\dfrac{1,6-1,3}{0,8}=0,375M$

$\rm d)$

$\rm NaOH\ dư\ nên\ kết\ tủa\ chỉ\ có\ Mg(OH)_2$

$\rm MgCl_2+2NaOH\to Mg(OH)_2+2NaCl$

$\rm n_{Mg(OH)_2}=n_{MgCl_2}=0,2(mol)$

$\rm m_{Mg(OH)_2}=0,2.58=11,6(g)$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{16,1135}}{{24,79}} = 0,65\,mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
27a + 24b = 12,9\\
1,5a + b = 0,65
\end{array} \right.\\
 \Rightarrow a = 0,3;b = 0,2\\
{m_{Al}} = 0,3 \times 27 = 8,1g\\
{m_{Mg}} = 0,2 \times 24 = 4,8g\\
c)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,3\,mol\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2\,mol\\
{n_{HCl}} = 0,8 \times 2 - 0,65 \times 2 = 0,3\,mol\\
{C_M}MgC{l_2} = \dfrac{{0,2}}{{0,8}} = 0,25M\\
{C_M}AlC{l_3} = \dfrac{{0,3}}{{0,8}} = 0,375M\\
{C_M}HCl = \dfrac{{0,3}}{{0,8}} = 0,375M\\
b)\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
{n_{Mg{{(OH)}_2}}} = {n_{Mg}} = 0,2\,mol\\
{m_{Mg{{(OH)}_2}}} = 0,2 \times 58 = 11,6g
\end{array}\)