`Na_2CO_3 + 2HCl → 2NaCl + CO_2 + H_2O` 

Đặt `n_{NaCl} = x(mol)`

Theo pt: 

`n_{CO_2} = 1/2n_{NaCl}= 0,5x(mol)`

Ta có: 

`C%_{NaCl} = {58,5x}/{200 + 120 - 44. 0,5x} .100 = 20%`

`→x = 1,017(mol)`

Theo pt: 

`n_{HCl} = n_{NaCl} = 1,017(mol)`

`C%_{HCl} = {1,017 . 36,5}/{120} .100 = 30,93%`

`n_{Na_2CO_3} = n_{CO_2} = 0,5x = 0,5085(mol)`

`C%_{Na_2CO_3} = {0,5085 . 106}/{200} .100 = 26,95%`

\(\begin{array}{l}
Dat\,n_{Na_2CO_3}=x(mol)\\
Na_2CO_3+2HCl\to 2NaCl+CO_2+H_2O\\
Theo\,PT:\,n_{NaCl}=2x(mol);n_{CO_2}=x(mol)\\
m_{dd\,spu}=200+120-44x=320-44x(g)\\
\to C\%_{NaCl}=\frac{58,5.2x}{320-44x}.100\%=20\%\\
\to x\approx 0,5(mol)\\
Theo\,PT:\,n_{HCl}=2x=1(mol)\\
\to\begin{cases} C\%_{HCl}=\frac{1.36,5}{120}.100\%\approx 30,42\%\\ C\%_{Na_2CO_3}=\frac{0,5.106}{200}.100\%=26,5\% \end{cases}
\end{array}\)