a,

$n_{HCl}=\dfrac{73.25\%}{36,5}=0,5(mol)$

$n_{NaOH}=0,15(mol)$

$NaOH+HCl\to NaCl+H_2O$

$\Rightarrow n_{HCl\text{dư}}=n_{NaCl}=0,15(mol)$

$\Rightarrow n_{HCl\text{pứ}}=0,5-0,15=0,35(mol)$

$2X+2HCl\to 2XCl+2$

$\Rightarrow n_{X}=0,35(mol)$

$\overline{M}_X=\dfrac{5,65}{0,35}=16,14$

$\Rightarrow$ 2 kim loại là Li (x mol), Na (y mol)

b,

Ta có: $7x+23y=5,65$            

$n_{hh}=0,35(mol)\Rightarrow x+y=0,35$

Suy ra $x=0,15; y=0,2$

$n_{LiCl}=0,15(mol)\Rightarrow m_{LiCl}=6,375g$

$\sum m_{NaCl}=(0,2+0,15).58,5=20,475g$

Đáp án:

 a) $Li,Na$

b) 20,475g và 6,375g

Giải thích các bước giải:

 \(\begin{array}{l}
2X + 2HCl \to 2XCl + {H_2}(1)\\
HCl + NaOH \to NaCl + {H_2}O(2)\\
{n_{NaOH}} = 0,15 \times 1 = 0,15\,mol\\
{n_{HCl(2)}} = {n_{NaOH}} = 0,15\,mol\\
{n_{HCl}} = \dfrac{{73 \times 25\% }}{{36,5}} = 0,5\,mol\\
{n_{HCl(1)}} = 0,5 - 0,15 = 0,35\,mol\\
{n_X} = {n_{HCl(1)}} = 0,35mol\\
{M_X} = \dfrac{{5,65}}{{0,35}} = 16,14(g/mol)\\
 \Rightarrow hh\,X:Natri(Na),Liti(Li)\\
hh:Li(a\,mol),Na(b\,mol)\\
7a + 23b = 5,65\\
a + b = 0,35\\
 \Rightarrow a = 0,15;b = 0,2\\
{m_{NaCl}} = (0,2 + 0,15) \times 58,5 = 20,475g\\
{m_{LiCl}} = 0,15 \times 42,5 = 6,375g
\end{array}\)