Đáp án:
\(\begin{array}{l}
a)\\
\% {V_{{O_3}}} = 66,67\% \\
\% {V_{{O_2}}} = 33,33\% \\
b)\\
{d_{A/He}} = 10,67
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2KI + {O_3} + {H_2}O \to 2KOH + {I_2} + {O_2}\\
{n_{{I_2}}} = \dfrac{{25,4}}{{254}} = 0,1\,mol\\
{n_{{O_3}}} = {n_{{O_2}}} = {n_{{I_2}}} = 0,1\,mol\\
{n_{{O_2}}} = \dfrac{{3,36}}{{22,4}} - 0,1 = 0,05\,mol\\
\% {V_{{O_3}}} = \dfrac{{0,1}}{{0,1 + 0,05}} \times 100\% = 66,67\% \\
\% {V_{{O_2}}} = 100 - 66,67 = 33,33\% \\
b)\\
{m_A} = 0,1 \times 48 + 0,05 \times 32 = 6,4g\\
{M_A} = \dfrac{{6,4}}{{0,15}} = \dfrac{{128}}{3}(g/mol)\\
{d_{A/He}} = \dfrac{{128}}{3}:4 = 10,67
\end{array}\)