Đáp án đúng: B
Z gồm $\left\{ \begin{array}{l}{{C}_{n}}{{H}_{{2n+3}}}:a\,mol\\{{C}_{2}}{{H}_{5}}N{{O}_{2}}:\,b\,mol\\{{C}_{6}}{{H}_{{14}}}{{N}_{2}}{{O}_{2}}:\,c\,mol\end{array} \right.$
→${{n}_{{C{{O}_{2}}}}}=\,na+2b+\,6c$ và${{n}_{{{{N}_{2}}}}}=\,\frac{a}{2}+\,\frac{b}{2}+\,c$
→${{n}_{{C{{O}_{2}}}}}\,+\,{{n}_{{{{N}_{2}}}}}=\,na+\,\frac{a}{2}\,+\,\frac{{5b}}{2}+7c=\,0,81$ (1)
${{n}_{{{{H}_{2}}O}}}=\,\frac{{a(2n+3)}}{2}+\frac{5}{2}b+\,7c=\,0,91(2)$
(2) – (1) a = 0,1
${{n}_{Z}}=\,0,2\to $ b +c = 0,1 (3)
Bảo toàn O:
2(b+c) +2${{n}_{{{{O}_{2}}}}}$ =$2{{n}_{{C{{O}_{2}}}}}\,+\,{{n}_{{{{H}_{2}}O}}}\,$ →${{n}_{{C{{O}_{2}}}}}\,=\,0,68$
→${{n}_{{{{N}_{2}}}}}=\frac{a}{2}+\frac{b}{2}+\,c=\,0,81-0,68(4)$
(3), (4) → b = 0,04 và c = 0,06
${{n}_{{C{{O}_{2}}}}}=\,na\,+2b\,+\,6c\,=\,0,68$ → n = 2,4
→ X chưa ;${{C}_{3}}{{H}_{9}}:\,0,04$
Vậy Z chứa${{C}_{2}}{{H}_{7}}N:\,0,06\,$ ;${{C}_{3}}{{H}_{9}}N:\,0,04$ ;${{C}_{2}}{{H}_{5}}N{{O}_{2}}$ : 0,04;${{C}_{6}}{{H}_{{14}}}{{N}_{2}}{{O}_{2}}:\,0,06$
→ %${{C}_{2}}{{H}_{7}}N=$ 16,05%
