Đáp án:

\(CTHH:S{O_3}\)

Giải thích các bước giải:

\(\begin{array}{l}
CTHH:{S_x}{O_y}\\
{n_A} = \dfrac{V}{{22,4}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{M_A} = \dfrac{m}{n} = \dfrac{{20}}{{0,25}} = 80dvC\\
\% O = \dfrac{{y{M_O}}}{{{M_A}}} \times 100\%  = 60\% \\
 \Rightarrow y = \dfrac{{80 \times 0,6}}{{16}} = 3\\
{M_A} = x{M_S} + 3{M_O} = 80\\
 \Rightarrow x = \dfrac{{80 - 3 \times 16}}{{32}} = 1\\
 \Rightarrow CTHH:S{O_3}
\end{array}\)

$n_{A}=\dfrac{V_A}{22,4}=\dfrac{5,6}{22,4}=0,25(mol)$

$\leftrightarrow M_A=\dfrac{m_A}{n_A}=\dfrac{20}{0,25}=80(g/mol)$

$\%O=60\%$

$\to m_O=\dfrac{80.60}{100}=48(g)$

$\to n_O=\dfrac{48}{16}=3(mol)$

$\%S=100\%-60\%=40\%$

$\to m_S=\dfrac{80.40}{100}=32(g)$

$\to n_S=\dfrac{32}{32}=1(mol)$

Suy ra CTHH: $SO_3$