Đáp án + Giải thích các bước giải:

Bài 1:

$n_{C_2H_2}=\frac{3,36}{22,4}=0,15(mol)$

$C_2H_2+2Br_2\to C_2H_2Br_4$

Theo PT: $n_{Br_2}=2n_{C_2H_2}=0,3(mol)$

$\to V_{dd\,Br_2}=\frac{0,3}{2}=0,15(l)$

Bài 2:

$n_{C_2H_5OH}=\frac{4,6}{46}=0,1(mol)$

$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$

Theo PT: $n_{H_2}=\frac{1}{2}n_{C_2H_5OH}=0,05(mol)$

$\to V_{H_2(đktc)}=0,05.22,4=1,12(l)$

Em tham khảo nha:

\(\begin{array}{l}
1)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{n_{{C_2}{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{B{r_2}}} = 2{n_{{C_2}{H_2}}} = 0,3\,mol\\
{V_{{\rm{dd}}B{r_2}}} = \dfrac{{0,3}}{2} = 0,15l\\
2)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
{n_{{C_2}{H_5}OH}} = \dfrac{{4,6}}{{46}} = 0,1\,mol\\
{n_{{H_2}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{V_{{H_2}}} = 0,05 \times 22,4 = 1,12l
\end{array}\)