Đáp án:
\(\begin{array}{l}
a)\\
CTPT:{C_2}{H_4}O;{C_3}{H_6}O\\
b)\\
{m_{{C_2}{H_4}O}} = 0,88g\\
{m_{{C_3}{H_6}O}} = 0,58g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_n}{H_{2n}}O + \dfrac{{3n - 1}}{2}{O_2} \to nC{O_2} + n{H_2}O\\
{n_{C{O_2}}} = \dfrac{{1,568}}{{22,4}} = 0,07\,mol \Rightarrow {n_{{C_n}{H_{2n}}O}} = \dfrac{{0,07}}{n}\,mol\\
{M_{{C_n}{H_{2n}}O}} = \dfrac{{1,46}}{{\dfrac{{0,07}}{n}}} = \dfrac{{146n}}{7}\,g/mol\\
\Rightarrow 14n + 16 = \dfrac{{146n}}{7} \Rightarrow n = 2,33\\
\Rightarrow CTPT:{C_2}{H_4}O;{C_3}{H_6}O\\
b)\\
hh:{C_2}{H_4}O(a\,mol);{C_3}{H_6}O(b\,mol)\\
\left\{ \begin{array}{l}
44a + 58b = 1,46\\
2a + 3b = 0,07
\end{array} \right.\\
\Rightarrow a = 0,02;b = 0,01\\
{m_{{C_2}{H_4}O}} = 0,02 \times 44 = 0,88g\\
{m_{{C_3}{H_6}O}} = 1,46 - 0,88 = 0,58g
\end{array}\)
