Đáp án:
34,46
Giải thích các bước giải:
${n_A} = \dfrac{{17,92}}{{22,4}} = 0,8mol$
$\begin{array}{*{20}{c}}
{S{O_2}:64}&{}&{}&{}&{24} \\
{}& \searrow &{}& \nearrow &{} \\
{}&{}&{56}&{}&{} \\
{}& \nearrow &{}& \searrow &{} \\
{{O_2}:32}&{}&{}&{}&8
\end{array} \Rightarrow \dfrac{{{n_{S{O_2}}}}}{{{n_{{O_2}}}}} = \dfrac{3}{1}$
$ \Rightarrow {n_{S{O_2}}} = 0,6mol;{n_{{O_2}}} = 0,2mol$
$2S{O_2} + {O_2}\overset {{V_2}{O_5}} \leftrightarrows 2S{O_3}$
H = 75%
$ \Rightarrow {n_{{O_2}pu}} = 0,2.\dfrac{{75}}{{100}} = 0,15mol;{n_{S{O_3}}} = {n_{S{O_2}pu}} = 2{n_{{O_2}pu}} = 0,3mol$
⇒ Hỗn hợp B gồm: ${O_2}(0,05{\text{ mol); S}}{{\text{O}}_2}(0,3{\text{ mol); S}}{{\text{O}}_3}(0,3{\text{ mol)}}$
$\begin{gathered}
\Rightarrow {M_B} = \dfrac{{\sum m }}{{\sum n }} = \dfrac{{{m_{{O_2}}} + {m_{S{O_2}}} + {m_{S{O_3}}}}}{{{n_{{O_2}}} + {n_{S{O_2}}} + {n_{S{O_3}}}}} \hfill \\
= \frac{{32.0,05 + 64.0,3 + 0,3.80}}{{0,05 + 0,3 + 0,3}} = 68,92 \hfill \\
\Rightarrow {d_{B/{H_2}}} = \dfrac{{{M_B}}}{{{M_{{H_2}}}}} = 34,46 \hfill \\
\end{gathered} $
