a) $HCl + AgNO_3 \to AgCl + HNO_3$

$n_{HCl} = n_{AgNO_3} = n_{AgCl} = \dfrac{14,35}{143,5} = 0,1(mol)$

$V_{dd\ HCl} = \dfrac{0,1}{2} = 0,05(lít)$

Thiếu nồng độ dung dịch $AgNO_3$ nên không thể tính thể tích

b)

$n_{HNO_3} = n_{AgCl} = 0,1(mol)$
$C_{M_{HNO_3}} = \dfrac{0,1}{0,05 + V_{dd\ AgNO_3}}$

a)

\(n_{H_2SO_4}=0,5a\left(mol\right)\); n_KOH = 2.0,2 = 0,4 (mol)

\(n_{Al\left(OH\right)_3}=\dfrac{0,39}{78}=0,005\left(mol\right)\)

TH1: 

D chứa axit dư

PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_{\text{4}}+2H_2O\)

               0,4------->0,2

=> \(n_{H_2SO_4\left(D\right)}=0,5a-0,2\left(mol\right)\)

PTHH: \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

                0,005---->0,0075

=> 0,5a - 0,2 = 0,0075.2

=> a = 0,43

TH2: D chứa KOH dư

PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_{\text{4}}+2H_2O\)

                 a<-------0,5a

=> \(n_{KOH\left(D\right)}=0,4-a\left(mol\right)\)

PTHH: \(Al\left(OH\right)_3+KOH\rightarrow KAlO_2+2H_2O\)

              0,005---->0,005

=> 0,4 - a = 0,005.2

=> a = 0,39 

b)

Gọi số mol Fe₃O₄, FeCO₃ là x, y (mol)

=> 232x + 116y = 2,688 (1)

TH1: a = 0,43 

\(n_{H_2SO_4}=0,1.0,43=0,043\left(mol\right)\)

PTHH: \(Fe_3O_4+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+FeSO_4+4H_2O\)

             \(FeCO_3+H_2SO_4\rightarrow FeSO_4+CO_2+H_2O\)

=> 4x + y = 0,043 (2)

(1)(2) => \(\left\{{}\begin{matrix}x=\dfrac{23}{2320}\left(mol\right)\\y=\dfrac{97}{29000}\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{Fe_3O_4}=\dfrac{23}{2320}.232=2,3\left(g\right)\\m_{FeCO_3}=\dfrac{97}{29000}.116=0,388\left(g\right)\end{matrix}\right.\)

TH2: a = 0,39 

=> \(n_{H_2SO_4}=0,39.0,1=0,039\left(mol\right)\)

=> 4x + y = 0,039 (3)

(1)(3) => \(\left\{{}\begin{matrix}x=\dfrac{459}{58000}\left(mol\right)\\y=\dfrac{213}{29000}\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{Fe_3O_4}=\dfrac{459}{58000}.232=1,836\left(g\right)\\m_{FeCO_3}=\dfrac{213}{29000}.116=0,852\left(g\right)\end{matrix}\right.\)