Đáp án:
a) 24,67
b) $\begin{gathered}
\% {m_{{H_2}}} = 2,7\% ;\% {m_{{N_2}}} = 37,84\% \hfill \\
\% {m_{C{O_2}}} = 59,46\% \hfill \\
\end{gathered} $
c) $\% {V_{{N_2}}} = \% {V_{C{O_2}}} = 25\% ;\% {V_{{H_2}}} = 50\% $
Giải thích các bước giải:
a) ${V_{{H_2}}} = {V_{{N_2}}} = {V_{C{O_2}}} \Rightarrow {n_{{H_2}}} = {n_{{N_2}}} = {n_{C{O_2}}}$
Đặt số mol mỗi khí là 1 mol
$\overline {\overline M = \dfrac{{\sum m }}{{\sum n }} = \dfrac{{28 + 2 + 44}}{{1 + 1 + 1}} = 24,67} $
b) $\sum m = 28 + 44 + 2 = 74g$
$\begin{gathered}
\Rightarrow \% {m_{{H_2}}} = \dfrac{2}{{74}}.100\% = 2,7\% \hfill \\
\% {m_{{N_2}}} = \dfrac{{28}}{{74}}.100\% = 37,84\% \hfill \\
\% {m_{C{O_2}}} = 100 - 37,84 - 2,7 = 59,46\% \hfill \\
\end{gathered} $
c) ${M_Y} = 9,5.2 = 19$
$\begin{array}{*{20}{c}}
{X:24,67}&{}&{}&{}&{17} \\
{}& \searrow &{}& \nearrow &{} \\
{}&{}&{19}&{}&{} \\
{}& \nearrow &{}& \searrow &{} \\
{{H_2}:2}&{}&{}&{}&{5,67}
\end{array} \Rightarrow \dfrac{{{n_X}}}{{{n_{{H_2}}}}} = 3$
Đặt: $\begin{gathered}
{n_X} = 3mol;{n_{{H_2}}} = 1mol \hfill \\
\Rightarrow {n_{{N_2}}} = 1mol;{n_{C{O_2}}} = 1mol;\sum {{n_{{H_2}}} = 2mol} \hfill \\
\Rightarrow \% {V_{{N_2}}} = \% {V_{C{O_2}}} = \dfrac{1}{4}.100\% = 25\% \hfill \\
\% {V_{{H_2}}} = \dfrac{2}{4}.100\% = 50\% \hfill \\
\end{gathered} $
