a) Gọi \(n_{N_2\left(pư\right)}=a\left(l\right)\left(a\le5\right)\)
PTHH: \(N_2+3H_2\xrightarrow[]{t^o,p,xt}2NH_3\)
a---->3a---------->2a
`=>` \(B:\left\{{}\begin{matrix}N_2:5-a\left(l\right)\\H_2:10-3a\left(l\right)\\NH_3:2a\left(l\right)\end{matrix}\right.\)
`=>` \(5-a+10-3a+2a=10\)
`=> a = 2,5 (l)`
`=>` \(\left\{{}\begin{matrix}\%V_{N_2\left(dư\right)}=\dfrac{5-2,5}{10}.100\%=25\%\\\%V_{H_2\left(dư\right)}=\dfrac{10-3.2,5}{10}.100=25\%\\\%V_{NH_3}=100\%-25\%-25\%=50\%\end{matrix}\right.\)
b) Xét tỉ lệ: \(\dfrac{V_{N_2}}{1}>\dfrac{V_{H_2}}{3}\left(\dfrac{5}{1}>\dfrac{10}{3}\right)\Rightarrow\) Hiệu suất phản ứng tính theo `H_2`
`=>` \(H=\dfrac{3.2,5}{10}.100\%=75\%\)
c) Giá sử có 1 mol B, do các khí đo ở cùng đk
`=>` \(\left\{{}\begin{matrix}n_{N_2}=0,25\left(mol\right)\\n_{H_2}=0,25\left(mol\right)\\n_{NH_3}=0,5\left(mol\right)\end{matrix}\right.\)
`=>` \(M_B=\dfrac{0,25.28+0,25.2+0,5.17}{1}=16\left(g/mol\right)\)
`=>` \(d_{B/kk}=\dfrac{16}{29}\approx0,552\)
