\(AlX_3+3AgNO_3\rightarrow3AgX+Al\left(NO_3\right)_3\)
\(2AgX\rightarrow2Ag+X_2\)
Ta có :
\(n_{Ag}=\frac{32,4}{108}=0,3\left(mol\right)\)
\(\Rightarrow n_{AgX}=n_{Ag}=0,3\left(mol\right)\)
\(\Rightarrow n_{AgX3}=0,1\left(mol\right)\)
\(M_{AgX3}=\frac{13,35}{0,1}=133,5\left(\frac{g}{mol}\right)\)
\(\Rightarrow M_X=35,5\left(\frac{g}{mol}\right)\)
\(\Rightarrow X:Clo\left(Cl\right)\)