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 \[\begin{array}{l} \% {m_{N{a_2}S{O_3}}} = \frac{{0,1242.126}}{{28,56}}.100 = 54,8\% \\ \% {m_{NaHS{O_3}}} = \frac{{0,0108.104}}{{28,56}}.100 = 3,93\% \\ \% {m_{N{a_2}S{O_4}}} = 41,27\% \end{array}\]