Đáp án:
$C_3H_8O$
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_n}{H_{2n + 1}}OH + \dfrac{{3n}}{2}{O_2} \xrightarrow{t^0} nC{O_2} + (n + 1){H_2}O\\
{n_{C{O_2}}} = \dfrac{{9,9}}{{44}} = 0,225\,mol\\
{n_Y} = \dfrac{{0,225}}{n}\,mol \Rightarrow {M_Y} = \dfrac{{4,5}}{{\dfrac{{0,225}}{n}}} = 20n\,g/mol\\
\Rightarrow 14n + 18 = 20n \Rightarrow n = 3\\
CTPT:{C_3}{H_8}O\\
b)\\
C{H_3} - C{H_2} - C{H_2} - OH:propan - 1 - ol\\
c)\\
{C_3}{H_7}OH + CuO \to {C_2}{H_5}CHO + Cu + {H_2}O
\end{array}\)
