`#Ly`
Đáp án:
`V=6,135525` $(l)$
Lời giải:
`n_(NH_3` `=` `m/M` `=` `(11,22)/(17)=0,66` $(mol)$
PTHH: $N_2$ $+$ $3H_2$ $\xrightarrow[]{t^o}$ $2NH_3$
Theo PT, $n_{H_2(lt)}$ `=` `(3)/(2)n_(NH_3` $=$ `(3)/(2).0,66=0,99` $(mol)$
Do `H=25%` `=>` $n_{H_2(tt)}$ `=` `0,99.25%=0,2475` $(mol)$
`V_(H_2(đkc)` `=` `n.24,79=0,2475.24,79=6,135525` $(l)$