Đáp án:
b) Hỗn hợp A: 50% SO2; 50% O2
hỗn hợp B: 12,5% SO2; 37,5% O2; 50% SO3
c) 40% O2; 80% SO2
Giải thích các bước giải:
a) $2S{O_2} + {O_2}\overset {{V_2}{O_5},{t^o}} \leftrightarrows 2S{O_3}$
b) ${M_A} = 12.4 = 48;{M_B} = 3,75.16 = 60$
Hỗn hợp A:
$\begin{array}{*{20}{c}}
{S{O_2}:64}&{}&{}&{}&{16} \\
{}& \searrow &{}& \nearrow &{} \\
{}&{}&{48}&{}&{} \\
{}& \nearrow &{}& \searrow &{} \\
{{O_2}:32}&{}&{}&{}&{16}
\end{array} \Rightarrow \dfrac{{{n_{S{O_2}}}}}{{{n_{{O_2}}}}} = 1$
$ \Rightarrow \% {V_{S{O_2}}} = \% {V_{{O_2}}} = 50\% $
Đặt ${n_{S{O_2}}} = {n_{{O_2}}} = 1{\text{ mol}}$
PTHH: $2S{O_2} + {O_2}\overset {{V_2}{O_5},{t^o}} \leftrightarrows 2S{O_3}$
Bđ: 1 1
Pư: 2x x 2x
Sau: 1-2x 1-x 2x
${n_B} = 2 - x$
Bảo toàn khối lượng: ${m_A} = {m_B} \Leftrightarrow 2.48 = 60.(2 - x) \Rightarrow x = 0,4$
Hỗn hợp B gồm: $S{O_2}(0,2mol);{O_2}(0,6mol);S{O_3}(0,8mol)$
$\begin{gathered}
\% {V_{S{O_2}}} = \dfrac{{0,2}}{{1,6}}.100\% = 12,5\% ;\% {V_{{O_2}}} = \dfrac{{0,6}}{{1,6}}.100\% = 37,5\% \hfill \\
\% {V_{S{O_3}}} = 100 - 12,5 - 37,5 = 50\% \hfill \\
\end{gathered} $
c)
$\begin{gathered}
{m_{{O_2}pu}} = 0,4.32 = 12,8g;{m_{S{O_2}pu}} = 0,8.64 = 51,2g \hfill \\
\Rightarrow \% {m_{{O_2}pu}} = \dfrac{{12,8}}{{32}}.100\% = 40\% \hfill \\
\% {m_{S{O_2}pu}} = \dfrac{{51,2}}{{64}}.100\% = 80\% \hfill \\
\end{gathered} $
