Đáp án:
1) b) 66,67% ;33,33%
2) 80%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
b)\\
nB{r_2} = 0,2 \times 1 = 0,2\,mol\\
= > n{C_2}{H_2} = 0,1\,mol\\
\% V{C_2}{H_2} = \frac{{0,1 \times 22,4}}{{3,36}} \times 100\% = 66,67\% \\
\% VC{H_4} = 100 - 66,67 = 33,33\% \\Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}\\
nCa{C_2} = \frac{8}{{64}} = 0,125\,mol\\
= > n{C_2}{H_2}(lt) = 0,125\,mol\\
H = \frac{{0,1}}{{0,125}} \times 100\% = 80\%
\end{array}\)