\(n_{NH_3}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
\(n_{H_2}=\dfrac{1,5.3}{2}=2,25\left(mol\right)\)
\(n_{N_2}=\dfrac{1.5}{2}=0,75\left(mol\right)\)
\(A_{H_2}=2,25.6.10^{23}=13,5.10^{23}\left(phan.tu\right)\)
\(A_{N_2}=0,75.6.10^{23}=4,5.10^{23}\left(phan.tu\right)\)
\(V_{NH_3}=1,5.22,4=33,6\left(l\right)\)