N2+3H2\(\rightarrow\)2NH3
Gọi số mol N2=3 mol, số mol H2 là 7 mol
Tỉ lệ: \(\dfrac{3}{1}>\dfrac{7}{3}\)Tính các chất theo H2
\(n_{H_2\left(pu\right)}=\dfrac{25}{100}.7=1,75mol\)
\(n_{H_2\left(dư\right)}=7-1,75=5,25mol\)
\(n_{N_2\left(Pu\right)}=\dfrac{1}{3}n_{H_2}=\dfrac{1,75}{3}mol\)
\(n_{N_2\left(dư\right)}=3-\dfrac{1,75}{3}=\dfrac{7,25}{3}mol\)
\(n_{NH_3}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.1,75=\dfrac{7}{6}mol\)
\(\overline{M}=\dfrac{m_{hh}}{n_{hh}}=\dfrac{5,25.2+28.\dfrac{7,25}{3}+17\dfrac{7}{6}}{5,25+\dfrac{7,25}{3}+\dfrac{7}{6}}\approx11,09\)
Đáp án D