a) \(V_{O_2}=n_{O_2}.22,4=0,5.22,4=11,2\left(l\right)\)
b) \(n_{Al_2O_3}=\frac{m_{Al_2O_3}}{M_{Al_2O_3}}=\frac{10,2}{102}=0,1\left(mol\right)\)
c) \(n_{NH_3}=\frac{V_{NH_3}}{22,4}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(m_{NH_3}=n_{NH_3}.M_{NH_3}=0,6.17=10,2\left(g\right)\)