Đáp án:

\(\begin{array}{l}
\% {m_{NaCl}} = 20,56\% \\
\% {m_{NaBr}} = 47,52\% \\
\% {m_{NaI}} = 31,92\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
hh:NaCl(a\,mol),NaBr(b\,mol),NaI(c\,mol)\\
58,5a + 103b + 150c = m(1)\\
T{N_1}:\\
2NaBr + C{l_2} \to 2NaCl + B{r_2}\\
2NaI + C{l_2} \to 2NaCl + {I_2}\\
hh\,Y:NaCl\\
 \Rightarrow 58,5a + 58,5b + 58,5c = 60\%  \times m(2)\\
T{N_2}:\\
2NaI + B{r_2} \to 2NaBr + {I_2}\\
 \Rightarrow hh\,Z:NaCl,NaBr\\
58,5a + 103b + 103b = 150\%  \times 60\%  \times m(3)\\
(1) - (3) \Rightarrow 47c = 0,1m \Rightarrow c = \dfrac{{0,1m}}{{47}} = \dfrac{m}{{470}}(mol)\\
(1) - (2) \Rightarrow 44,5b + 91,5c = 0,4m\\
 \Rightarrow b = \dfrac{{193m}}{{41830}}(mol)\\
a = \dfrac{{0,6m}}{{58,5}} - \dfrac{{193m}}{{41830}} - \dfrac{m}{{470}} = \dfrac{{61m}}{{17355}}(mol)\\
\% {m_{NaCl}} = \dfrac{{\dfrac{{61m}}{{17355}} \times 58,5}}{m} \times 100\%  = 20,56\% \\
\% {m_{NaBr}} = \dfrac{{\dfrac{{193m}}{{41830}} \times 103}}{m} \times 100\%  = 47,52\% \\
\% {m_{NaI}} = 100 - 20,56 - 47,52 = 31,92\% 
\end{array}\)