Đáp án:
1) 3,52 g
Giải thích các bước giải:
1)
${C_2}{H_5}OH + CuO \to C{H_3}CHO + Cu + {H_2}O$
$\begin{gathered}
{n_{{C_2}{H_5}OH}} = 0,1mol \hfill \\
H = 80\% \Rightarrow {n_{{C_2}{H_5}OH(pu)}} = 0,1.H = 0,08mol \hfill \\
\Rightarrow {n_{C{H_3}CHO}} = 0,08.44 = 3,52g \hfill \\
\end{gathered} $
2) Gọi CTPT của anken là ${C_n}{H_{2n}}$
$\begin{gathered}
{n_{hh}} = \dfrac{V}{{22,4}};{n_{C{H_4}}} = \dfrac{b}{{22,4}} \hfill \\
\Rightarrow {n_{{C_n}{H_{2n}}}} = \dfrac{{V - b}}{{22,4}} \hfill \\
{m_{binh\tan g}} = {m_{{C_n}{H_{2n}}}} = a \hfill \\
\Rightarrow a = 14n.\dfrac{{V - b}}{{22,4}} \hfill \\
\Rightarrow n = \dfrac{{22,4a}}{{14(V - b)}} \hfill \\
\end{gathered} $
3)
$\begin{gathered}
{n_{anken}} = \dfrac{V}{{22,4}} \hfill \\
\Rightarrow {n_{B{r_2}}} = {n_{anken}} = \dfrac{V}{{22,4}} \hfill \\
\Rightarrow {m_{B{r_2}}} = \dfrac{V}{{22,4}}.160 \hfill \\
\end{gathered} $
