\(\begin{array}{l}
200ml=0,2l\\
n_{Br_2}=0,2.1,5=0,3(mol)\\
SO_2+Br_2+2H_2O\to 2HBr+H_2SO_4\\
Theo\,PT:\,n_{SO_2}=n_{Br_2}=0,3(mol)\\
\text{Bảo toàn S: }n_{Na_2SO_3}+n_{NaHSO_3}=n_{SO_2}=0,3(mol)\\
\to n_{Na_2SO_3}=n_{NaHSO_3}=0,15(mol)\\
\to m_X=0,15(126+104)=34,5(g)\\
\to\begin{cases} V=0,3.22,4=6,72(l)\\ \%m_{Na_2SO_3}=\frac{0,15.126}{34,5}.100\%\approx 54,78\%\\ \%m_{NaHSO_3}=100-36,5=45,22\% \end{cases}
\end{array}\)

Đáp án:

\(V = 3,36{\text{ lít}}\)

\(\% {m_{N{a_2}S{O_3}}} = 54,78\% ;\% {m_{NaH{\text{S}}{O_4}}} = 45,22\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(N{a_2}S{O_3} + 2HCl\xrightarrow{{}}2NaCl + S{O_2} + {H_2}O\)

\(NaH{\text{S}}{O_3} + HCl\xrightarrow{{}}NaCl + S{O_2} + {H_2}O\)

\(B{{\text{r}}_2} + S{O_2} + 2{H_2}O\xrightarrow{{}}{H_2}S{O_4} + 2HB{\text{r}}\)

Gọi số mol \(Na_2SO_3;NaHSO_3\) đều là \(x\)

\( \to {n_{B{{\text{r}}_2}}} = 0,2.1,5 = 0,3{\text{ mol = }}{{\text{n}}_{S{O_2}}}\)

\( \to {n_{S{O_2}}} = {n_{N{a_2}S{O_3}}} + {n_{NaH{\text{S}}{O_3}}} = x + x = 2{\text{x}} = 0,3 \to x = 0,15{\text{ mol}}\)

\( \to V = {V_{S{O_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)

\( \to {m_{N{a_2}S{O_3}}} = 0,15.(23.2 + 32 + 16.3) = 18,9{\text{ gam}}\)

\({m_{NaHS{O_3}}} = 0,15.104 = 15,6{\text{ gam}}\)

\( \to \% {m_{N{a_2}S{O_3}}} = \frac{{18,9}}{{18,9 + 15,6}} = 54,78\%  \to \% {m_{NaH{\text{S}}{O_4}}} = 45,22\% \)