Giải bài 8 trang 204 SGK Hóa học 11

a)$\begin{align} & 2C{{H}_{2}}=C{{H}_{2}}+{{O}_{2}}\xrightarrow[xt]{{{t}^{o}}}2C{{H}_{3}}-CHO \\ & C{{H}_{3}}-CHO+2AgN{{O}_{3}}+3N{{H}_{3}}+{{H}_{2}}O\xrightarrow{{{t}^{o}}}C{{H}_{3}}C\text{OO}N{{H}_{4}}+2Ag\downarrow +2N{{H}_{4}}N{{O}_{3}} \\ \end{align}$

$\begin{align} & b){{n}_{Ag}}=\frac{16,2}{108}=0,15mol \\ & {{n}_{C{{H}_{3}}CHO\,tham\,gia}}=\frac{0,15}{2}=0,075mol \\ & {{n}_{{{C}_{2}}{{H}_{4}}\,dư}}=\frac{2,24}{22,4}-0,075=0,025mol \\ & {{n}_{etilen}}=0,075+0,025=0,1mol \\ \end{align}$

$H=\frac{0,075.100}{0,1}=75\%$