Đáp án:
1. a) 5,432 lít; 97% CH4; 3% C2H2
b) 2,595 g
c) 25,75 g
2. C2H2
Giải thích các bước giải:
1.
a) ${n_{hhkhi}} = \dfrac{{5,6}}{{22,4}} = 0,25mol;{n_{B{r_2}}} = 2.0,0075 = 0,015$
${C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}$
$\begin{gathered} \Rightarrow {n_{{C_2}{H_2}}} = \dfrac{1}{2}{n_{B{r_2}}} = 7,{5.10^{ - 3}}mol \hfill \\ \Rightarrow {n_{C{H_4}}} = 0,25 - 7,{5.10^{ - 3}} = 0,2425mol \Rightarrow {V_{C{H_4}}} = 5,432l \hfill \\ \% {V_{C{H_4}}} = \dfrac{{5,432}}{{5,6}}.100\% = 97\% \Rightarrow \% {V_{{C_2}{H_2}}} = 3\% \hfill \\ \end{gathered} $
b) ${n_{{C_2}{H_2}B{r_4}}} = {n_{{C_2}{H_2}}} = 7,{5.10^{ - 3}}mol \Rightarrow {m_{{C_2}{H_2}B{r_4}}} = 2,595g$
c) Bảo toàn nguyên tố C:
$\begin{gathered}
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 2{n_{{C_2}{H_2}}} + {n_{C{H_4}}} = 2.7,{5.10^{ - 3}} + 0,2425 = 0,2575mol \hfill \\
\Rightarrow {m_{CaC{O_3}}} = 25,75g \hfill \\
\end{gathered} $
2.
${M_A} = 13.2 = 26 \Rightarrow {n_A} = \dfrac{{5,2}}{{26}} = 0,2mol$
${n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2mol$
Gọi CTPT của A là ${C_x}{H_y}$
$ \Rightarrow y = \dfrac{{2{n_{{H_2}O}}}}{{{n_A}}} = 2 \Rightarrow x = \dfrac{{26 - 2}}{{12}} = 2$
Vậy A là ${C_2}{H_2}$