Câu trả lời:
\(2NaBr+Cl_2-->2NaCl+Br_2\left(1\right)\)
0,08________________0,08
\(2NaI+Cl_2-->2NaCl+I_2\left(2\right)\)
0,03_______________0,03
\(n_{NaBr}=\frac{8,24}{103}=0,08\left(mol\right)\)
\(n_{NaI}=\frac{4,5}{150}=0,03\left(mol\right)\)
=>\(m_{muoi}=m_{NaF}+m_{NaCl}=2,52+0,11.58,5=8,955\left(g\right)\)