Giải thích các bước giải:

 TN1:

$m_{dd\ Na_2CO_3}=63,6.1,25=79,5\ gam\Rightarrow n_{Na_2CO_3}=\frac{79,5.20\%}{106}=0,15\ mol$

Gọi số mol $H^+$ TN1 là a

$2H^++CO_3^{2^-}\to CO_2+H_2O$

Suy ra $n_{CO_3^{2^-}}=\frac{n_{H^+}}2\Rightarrow n_{H^+}=2n_{CO_3^{2^-}}=0,15.2=0,3\ mol$

$\Rightarrow n_{HCl}+n_{HNO_3}=0,3\ mol$

$n_{CO_2}=n_{CO_3^{2^-}}=0,15\ mol\Rightarrow V=0,15.22,4=3,36\ l$
TN2:

50 gam dd A chứa 0,15 mol $H^+$

$Cl^-+Ag^+\to AgCl$

$n_{AgCl}=\frac{14,35}{143,5}=0,1\ mol\Rightarrow n_{HCl}=n_{Cl^-}=0,1\ mol$

$\Rightarrow n_{HNO_3}=0,15-0,1=0,05\ mol$

Suy ra ở TN1 có 0,1 mol $HNO_3$ và 0,2 mol HCl

Trong 100 gam dung dịch A chứa 0,1 mol $HNO_3$ và 0,2 mol HCl

Suy ra $C\%_{HNO_3}=\frac{0,1.63}{100}\cdot 100\%=6,3\%;\text C\%_{HCl}=\frac{0,2.36,5}{100}\cdot 100\%=7,3\%%$

Dung dịch X thu được chứa $NaCl, NaNO_3$

Theo BTKL: $m_{dd\ A}+m_{dd\ Na_2CO_3}=m_{dd\ spu}+m_{CO_2}$

$\Rightarrow m_{dd\ spu}=100+79,5-0,15.44=172,9\ \text{gam}$

Ta có: $n_{NaCl}=n_{HCl}=0,2\ mol\Rightarrow C\%_{NaCl}=\frac{0,2.58,5}{172,9}\cdot 100\%=6,76\%$

$n_{NaNO_3}=n_{HNO_3}=0,1\ mol\Rightarrow C\%_{NaNO_3}=\frac{0,1.85}{172,9}\cdot 100\%=4,91\%$

- TN2:

$AgNO_3+HCl\to AgCl+HNO_3$

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

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

$\to 100g$ $A$ có $0,1.2=0,2$ mol $HCl$

- TN1:

$m_{dd Na_2CO_3}=63,6.1,25=79,5g$

$\to n_{Na_2CO_3}=\dfrac{79,5.20\%}{106}=0,15(mol)$

$2HCl+Na_2CO_3\to 2NaCl+CO_2+H_2O$

$2HNO_3+Na_2CO_3\to 2NaNO_3+CO_2+H_2O$

Ta có: $2n_{Na_2CO_3}=n_{HCl}+n_{HNO_3}$

$\to n_{HNO_3}=0,1(mol)$

$\to n_{NaCl}=n_{HCl}=0,2(mol); n_{NaNO_3}=n_{HNO_3}=0,1(mol)$

$n_{CO_2}=n_{Na_2CO_3}=0,15(mol)$

$\to m_{dd\text{spu}}=100+79,5-0,15.44=172,9g$

a,  

$V=0,15.22,4=3,36l$

b,

$C\%_{HCl}=\dfrac{0,2.36,5.100}{100}=7,3\%$

$C\%_{HNO_3}=\dfrac{0,1.63.100}{100}=6,3\%$

c,

$C\%_{NaCl}=\dfrac{0,2.58,5.100}{172,9}=6,77\%$

$C\%_{NaNO_3}=\dfrac{0,1.85.100}{172,9}=4,92\%$