PT: \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)
\(AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\)
Ta có: \(n_{AgNO_3}=0,1.0,6=0,06\left(mol\right)\)
Theo PT: \(n_{Al}=n_{AlCl_3}=\dfrac{1}{3}n_{AgNO_3}=0,02\left(mol\right)\)
⇒ m = mAl = 0,02.27 = 0,54 (g)
\(C_{M_{AlCl_3}}=\dfrac{0,02}{0,15}=\dfrac{2}{15}\left(M\right)\)