4Al + 3O2 →to 2Al2O3
a) \(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3\left(mol\right)\)
Theo PT: \(n_{Al}pư=2n_{Al_2O_3}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow m_{Al}pư=0,6\times27=16,2\left(g\right)\)
b) Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=\dfrac{3}{2}\times0,3=0,45\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,45\times22,4=10,08\left(l\right)\)