Đáp án:
% theo V: 10% N2; 30% H2; 60% NH3
% theo m: 20,6% N2; 4,4% H2; 75% NH3
Giải thích các bước giải:
${M_A} = 0,425.32 = 13,6$
Đặt: ${n_{{N_2}}} = x;{n_{{H_2}}} = 3x;{n_{N{H_3}}} = y$
${M_A} = \dfrac{{{m_{{H_2}}} + {m_{{N_2}}} + {m_{N{H_3}}}}}{{{n_{{H_2}}} + {n_{{N_2}}} + {n_{N{H_3}}}}} = \dfrac{{2.3x + 28x + 17y}}{{3x + x + y}} = \dfrac{{34x + 17y}}{{4x + y}} = 13,6$
$\begin{gathered}
\Rightarrow 34x + 17y = 54,4x + 13,6y \hfill \\
\Rightarrow 20,4x = 3,4y \Rightarrow \dfrac{x}{y} = \dfrac{1}{6} \hfill \\
\end{gathered} $
$ \Rightarrow {n_{{N_2}}} = 1mol;{n_{{H_2}}} = 3mol;{n_{N{H_3}}} = 6mol$
$\begin{gathered}
\% {V_{{N_2}}} = \dfrac{1}{{10}}.100\% = 10\% ;\% {V_{{H_2}}} = \dfrac{3}{{10}}.100\% = 30\% \hfill \\
\% {V_{N{H_3}}} = 100 - 10 - 30 = 60\% \hfill \\
\end{gathered} $
${m_A} = 13,6.10 = 136\left( g \right)$
$\begin{gathered}
\% {m_{{N_2}}} = \dfrac{{28.1}}{{136}}.100\% = 20,6\% ;\% {m_{{H_2}}} = \dfrac{{2.3}}{{136}} = 4,4\% \hfill \\
\% {m_{N{H_3}}} = 100 - 20,6 - 4,4 = 75\% \hfill \\
\end{gathered} $
