Có: \(\left\{{}\begin{matrix}n_{N_2}+n_{H_2}=\dfrac{11,2}{22,4}=0,5\\\dfrac{n_{N_2}}{n_{H_2}}=\dfrac{1}{4}\end{matrix}\right.\)
=> nN2 = 0,1 (mol); nH2 = 0,4 (mol)
PTHH: \(N_2+3H_2\underrightarrow{t^o,p,xt}2NH_3\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\) => N2 hết, H2 dư
PTHH: \(N_2+3H_2\underrightarrow{t^o,p,xt}2NH_3\)
______0,1---------------->0,2
=> VNH3 = 0,2.22,4 = 4,48 (l)