a) \(Ag^++X^-\rightarrow AgX\downarrow\)
Ta có \(n_{AgX}=n_{X^-}=n_{NaX}=\dfrac{11,7}{23+X}\)
\(\Rightarrow m_{AgX}=\dfrac{11,7}{23+X}.\left(108+X\right)=28,7\)
\(\Rightarrow X=35,5\) (Cl)
b) \(n_{Cl_2}=\dfrac{8,52}{71}=0,12\) mol
=> V = 22,4.0,12 = 2,688 lít