Gọi a là mol N2, 4a là mol H2 ban đầu
\(N_2+3H_2⇌2NH_3\)
Mà H = 25%
\(\Rightarrow n_{N2\left(pư\right)}=0,25a\left(mol\right)\)
\(n_{NH3}=0,5a\left(mol\right)\)
\(n_{NH3}=0,1\left(mol\right)\)
\(\Rightarrow0,5a=0,1\Leftrightarrow a=0,2\)
\(V_{N2}=0,2.22,4=4,48\left(l\right)\)
\(V_{H2}=0,2.4.22,4=17,92\left(l\right)\)