Đáp án:
A: $\% {V_{CO}} = \% {V_{{O_2}}} = 50\% $
B: $\% {V_{CO}} = \% {V_{{O_2}}} = 37,5\% ;\% {V_{C{O_2}}} = 25\% $
H = 40%
Giải thích các bước giải:
${M_A} = 1,875.16 = 30$
$\begin{array}{*{20}{c}}
{CO:28}&{}&{}&{}&2 \\
{}& \searrow &{}& \nearrow &{} \\
{}&{}&{30}&{}&{} \\
{}& \nearrow &{}& \searrow &{} \\
{{O_2}:32}&{}&{}&{}&2
\end{array} \Rightarrow \dfrac{{{n_{CO}}}}{{{n_{{O_2}}}}} = \dfrac{1}{1}$
$ \Rightarrow \% {V_{CO}} = \% {V_{{O_2}}} = 50\% $
${M_B} = 18,75.2 = 37,5$
PTHH: $CO + {O_2} \to C{O_2}$
Bđ: 1 1
Pư: x x x
Sau: 1-x 1-x x
Bảo toàn khối lượng: ${m_A} = {m_B} \Leftrightarrow 2.30 = (2 - x).37,5$
$\begin{gathered}
\Rightarrow x = 0,4 \Rightarrow {n_{CO}} = {n_{{O_2}}} = 0,6;{n_{C{O_2}}} = 0,4 \hfill \\
\Rightarrow \% {V_{CO}} = \% {V_{{O_2}}} = \dfrac{{0,6}}{{1,6}}.100\% = 37,5\% \hfill \\
\Rightarrow \% {V_{C{O_2}}} = \dfrac{{0,4}}{{1,6}}.100\% = 25\% \hfill \\
\end{gathered} $
$H = \dfrac{x}{1}.100\% = 40\% $
