\(NaX+AgNO3--->AgX+NaNO3\)
\(n_{NaX}=\frac{15,45}{23+X}\left(mol\right)\)
\(n_{AgX}=\frac{28,2}{108+X}\left(mol\right)\)
Theo pthh
\(n_{NaX}=n_{AgX}\Rightarrow\frac{15,45}{23+X}=\frac{28,2}{108+X}\)
\(\Rightarrow1668,6+15,45X=648,6+28,2X\)
\(\Rightarrow12,75X=1020\)
\(\Rightarrow X=80\left(Br\right)\)
Vậy CTHH: NaBr