Đáp án:
Giải thích các bước giải:
Lần lượt gọi số mol Na2SO3,NaHSO3, Na2SO4 là a,b,c
\[\begin{array}{l}
PTHH:\\
N{a_2}S{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + S{O_2} + {H_2}O\\
a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\
2NaHS{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + 2S{O_2} + 2{H_2}O\\
\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\\
\,\,\,\,\,S{O_2}\,\,\,\,\,\, + \,\,\,\,B{r_2} + \,\,\,\,2{H_2}O \to 2HBr + {H_2}S{O_4}\\
a + b\,\,\,\,\,\, \to \,a + b\\
a + b = {n_{B{r_2}}} = 0,135\\
126a + 104b + 142c = 28,56
\end{array}\]
7,14gA =1/4 khối lượng hỗn hợp ban đầu
\[\begin{array}{l}
2NaHS{O_3} + 2KOH \to {K_2}S{O_3} + N{a_2}S{O_3} + 2{H_2}O\\
\frac{1}{4}b\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\frac{1}{4}b\\
\Rightarrow \frac{1}{4}b = {n_{KOH}} = 2,{7.10^{ - 3}} \to b = 0,0108(3)
\end{array}\]
Từ 1,2,3 suy ra
a=0,1242
b=0,0108
c=0,083
\[\begin{array}{l}
\% {m_{N{a_2}S{O_3}}} = \frac{{0,1242.126}}{{28,56}}.100 = 54,8\% \\
\% {m_{NaHS{O_3}}} = \frac{{0,0108.104}}{{28,56}}.100 = 3,93\% \\
\% {m_{N{a_2}S{O_4}}} = 41,27\%
\end{array}\]
