PTHH:
\(N_{2}+3H_{2}\overset{xt,t^{o}, p}{\rightarrow}2NH_{3}\)
\(n_{NH_{3}}=1(mol)\)
Theo PTHH:
\(n_{N_{2}}=\frac{1}{2}n_{NH_{3}}=0,5(mol)\)
\(n_{H_{2}}=\frac{3}{2}n_{NH_{3}}=1,5(mol)\)
Hiệu suất 25% nên:
\(n_{N_{2}}=0,5.\frac{100}{25}=2(mol)\)
\(n_{H_{2}}=1,5.\frac{100}{25}=6(mol)\)
Vậy \(V_{N_{2}}=44,8l\) và \(V_{H_{2}}=134,4l\)