Em tham khảo nha:
\(\begin{array}{l}
22)\\
hh\,A:NaCl(a\,mol),NaBr(b\,mol),NaI(c\,mol)\\
58,5a + 103b + 150c = 1,732(1)\\
TN2:\\
2NaI + B{r_2} \to 2NaBr + {I_2}\\
58,5a + 103b + 103c = 1,685(2)\\
TN3:\\
2NaBr + C{l_2} \to 2NaCl + B{r_2}\\
2NaI + C{l_2} \to 2NaCl + {I_2}\\
58,5(a + b + c) = 1,4625(3)\\
(1),(2),(3) \Rightarrow a = 0,02;b = 0,004;c = 0,001\\
{C_M}NaCl = \dfrac{{0,02}}{{0,02}} = 1M\\
{C_M}NaBr = \dfrac{{0,004}}{{0,02}} = 0,2M\\
{C_M}NaI = \dfrac{{0,001}}{{0,02}} = 0,05M\\
23)\\
NaM + AgN{O_3} \to AgM + NaN{O_3}\\
nNaM = nAgM \Rightarrow \frac{{31,84}}{{23 + MM}} = \dfrac{{57,34}}{{108 + MM}}\\
\Rightarrow MM = 83,1333(g/mol)\\
\Rightarrow CTHH:NaBr,NaI
\end{array}\)
