Gọi \(\left\{{}\begin{matrix}n_{NaBr}=x\left(mol\right)\\n_{NaI}=y\left(mol\right)\end{matrix}\right.\)
\(2NaI+Br_2\rightarrow2NaBr+I_2\)
y y
\(\Rightarrow m_{muối\downarrow}=127y-80y=7,05\Rightarrow y=0,15mol\)
\(2NaBr_2+Cl_2\rightarrow2NaCl+Br_2\)
\(2NaI+Cl_2\rightarrow2NaCl+I_2\)
\(\Rightarrow m_{giảm}=\left(80x+127y\right)-35,5\left(x+y\right)=22,625\)
Thay \(y=0,15mol\) ta được: \(x=0,2mol\)
\(\%m_{NaI}=\dfrac{0,15\cdot150}{0,15\cdot150+0,2\cdot103}\cdot100\%=52,2\%\)
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