Đáp án:
\(\begin{array}{l}
a = 7,1875;b = 31,25\\
CTCT:C{H_3} - O - C{H_3}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_6}O + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 3{H_2}O\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
CaC{O_3} \xrightarrow{t^0} CaO + C{O_2}\\
{n_{C{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
H = 80\% \Rightarrow {n_{CaC{O_3}}} = \dfrac{{0,25 \times 100}}{{80}} = 0,3125\,mol\\
{m_{CaC{O_3}}} = 0,3125 \times 100 = 31,25g\\
{n_{{C_2}{H_6}O}} = \dfrac{{0,3125}}{2} = 0,15625\,mol\\
{m_{{C_2}{H_6}O}} = 0,15625 \times 46 = 7,1875g\\
CTCT:C{H_3} - O - C{H_3}
\end{array}\)