Đáp án:
1,D
2,A
Giải thích các bước giải:
1,
\(\begin{array}{l}
{N_2} + 3{H_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2N{H_3}\\
{n_{{N_2}}} = \dfrac{{15}}{{56}}mol\\
\to {n_{N{H_3}}} = 2{n_{{N_2}}} = \dfrac{{15}}{{28}}mol\\
\to {V_{N{H_3}}} = 12l\\
\to {V_{N{H_3}}}thực= \dfrac{{12 \times 25}}{{100}} = 3l
\end{array}\)
3,
Theo sơ đồ đường chéo ta có:
\(\dfrac{{{n_{N{O_2}}}}}{{{n_{NO}}}} = \dfrac{{19,5 \times 2 - 30}}{{46 - 19,5 \times 2}} = \dfrac{9}{7}\)
Mặt khác ta có: \({n_{N{O_2}}} + {n_{NO}} = 0,64mol\)
Giải hệ phương trình ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{n_{N{O_2}}} + {n_{NO}} = 0,64\\
\dfrac{{{n_{N{O_2}}}}}{{{n_{NO}}}} = \dfrac{9}{7}
\end{array} \right.\\
\to {n_{N{O_2}}} = 0,36mol \to {n_{NO}} = 0,28mol
\end{array}\)
\(\begin{array}{l}
NO + \dfrac{1}{2}{O_2} \to N{O_2}\\
2N{O_2} + \dfrac{1}{2}{O_2} + {H_2}O \to 2HN{O_3}\\
{n_{N{O_2}}} = {n_{NO}} = 0,28mol\\
\to {n_{{O_2}}}(NO) = \dfrac{1}{2}{n_{NO}} = 0,14mol\\
\to {n_{{O_2}}}(N{O_2}) = \dfrac{1}{4}{n_{N{O_2}}} = \dfrac{1}{4} \times (0,36 + 0,28) = 0,16mol\\
\to {n_{{O_2}}} = 0,14 + 0,16 = 0,3mol\\
\to {V_{{O_2}}} = 6,72l
\end{array}\)