PTHH: NaX + AgNO3 --> NaNO3 + AgX
2AgX --as--> 2Ag + X2
\(n_{Ag}=\dfrac{1,08}{108}=0,01\left(mol\right)\)
=> \(n_{NaX}=0,01\left(mol\right)\)
=> \(M_{NaX}=\dfrac{1,03}{0,01}=103\left(g/mol\right)\)
=> MX = 80 (g/mol)
=> X là Br
Muối A là NaBr (Natri Bromua)