Cho 11,2g nitơ tác dụng với 3g hiđrô thu được 38,08(l) hỗn hợp khí (ĐKTC) tính hiệu suất phản ứng

Đáp án:

 $H=25\%$

Giải thích các bước giải:

$ {n_{{N_2}}} = \frac{{11,2}}{{28}} = 0,4\,\,mol,\,\,{n_{{H_2}}} = \frac{3}{2} = 1,5\,\,mol$

$ {n_{\text {hh sau phản ứng}}} = \frac{{38,08}}{{22,4}} = 1,7\,\,mol$

Phương trình hóa học:

$\begin{array}{l} {N_2}{\rm{ }} + {\rm{ }}3{H_2}{\rm{ }} \to {\rm{ }}2N{H_3}\\ 0,4\,\,\,\,\,\,\,\,\,\,\,\,1,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \end{array}$

Nhận thấy:$\frac{{0,4}}{1} < \frac{{1,5}}{3} \to $ Hiệu suất phản ứng tính theo $N_2$

Gọi số mol của $N_2$ phản ứng là a mol

$\begin{array}{l} \,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{N_2}{\rm{ }} + {\rm{ }}3{H_2}{\rm{ }} \to {\rm{ }}2N{H_3}\\ Ban\,\,dau:\,\,\,\,0,4\,\,\,\,\,\,\,\,\,\,\,\,1,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\\ Pu\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,a\,\,\,\,\,\,\,\,\, \to \,\,\,\,3a \to \,\,\,\,\,\,\,\,\,\,\,\,2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\\ Sau\,\,pu\,\,\,\,\,\,\,:0,4 – a\,\,\,\,\,1,5 – 3a\,\,\,\,\,\,\,\,2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\\ \to {n_{hh\,\,sau\,\,pu}} = 0,4 – a + 1,5 – 3a + 2a = 1,7\,mol\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,a\,\,\,\, = 0,1\,\,\,\,mol\\ \to H = \frac{{{n_{{N_2}\,\,phan\,\,ung}}}}{{{n_{{N_2}\,\,ban\,\,dau}}}} \cdot 100\% = \frac{{0,1}}{{0,4}} \cdot 100\% = 25\% \end{array}$