a,

$n_{Na}=\dfrac{2,3}{23}=0,1(mol)$

$2Na+2HCl\to 2NaCl+H_2$

$\Rightarrow n_{H_2}=\dfrac{n_{Na}}{2}=0,05(mol)$

$\to V_2=V_{H_2}=0,05.22,4=1,12l$

b,

$n_{HCl}=n_{Na}=0,1(mol)$

$V_{dd HCl}=\dfrac{0,1}{1}=0,1l=100ml$

$\to V_1=100ml$

c,

$n_{NaCl}=n_{Na}=0,1(mol)$

$AgNO_3+NaCl\to AgCl+NaNO_3$

$\Rightarrow n_{AgCl\downarrow}=n_{NaCl}=0,1(mol)$

$\to a=0,1.143,5=14,35g$

$n_{Na}=\dfrac{2.3}{23}=0.1 \ (mol)\\\text{Phản ứng xảy ra:}\\2Na+2HCl\to 2NaCl+H_2\\ \ 0.1 \ \to \ \ 0.1 \xrightarrow{\quad \quad} 0.1 \xrightarrow{\quad}0.05 \ \ (mol)\\a,\\V_{1}=V_{HCl}=\dfrac{0.1}{1}=0.1 \ (l)\\b,\\V_2=V_{H_2}=0.05\times22.4=1.12 \ (l)\\\text{Dung dịch $A$ thu được là $NaCl$}\\\text{Cho dung dịch $A$ tác dụng với $AgNO_3$}\\NaCl+AgNO_3\to AgCl\downarrow+NaNO_3\\ \ \ 0.1 \ \ \ \ \ \ \xrightarrow{\quad \quad \quad} \ \ \ \ \ \ \ 0.1 \ \ (mol)\\c,\\a=m_{AgCl}=0.1\times143.5=14.35 \ (g)$