\(_{Al2\left(SO4\right)3}=\frac{68,4}{342}=0,2\left(mol\right)\)
Số NT có trong trong 68,4 gam Al2(SO4)3 là: \(0,2.6.10^{23}=1,2.10^{23}\)
\(n_{Al2\left(SO4\right)3}=\frac{68,4}{342}=0,2\left(mol\right)\)
\(\Rightarrow n_{Al}=0,4\left(mol\right)\Rightarrow SPT_{Al}=0,4.6.10^{23}=2,4.10^{23}\left(PT\right)\)
\(n_S=0,6\left(mol\right)\Rightarrow SPT_S=0,6.6.10^{23}=3,6.10^{23}\left(PT\right)\)
\(n_O=2,4\left(mol\right)\Rightarrow SPT_O=2,4.6.10^{24}=14,4.10^{23}\left(PT\right)\)