Đáp án:
b) 33,1 g
c) 16,56g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
2{C_3}{H_7}OH + 2Na \to 2{C_3}{H_7}ONa + {H_2}\\
n{H_2} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol \Rightarrow nA = 0,2 \times 2 = 0,4\,mol\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
nB{r_2} = 0,3 \times 1 = 0,3\,mol\\
n{C_6}{H_5}OH = \dfrac{{0,3}}{3} = 0,1\,mol\\
\Rightarrow n{C_3}{H_7}OH = 0,4 - 0,1 = 0,3\,mol\\
\% m{C_6}{H_5}OH = \dfrac{{0,1 \times 94}}{{0,1 \times 94 + 0,3 \times 60}} \times 100\% = 34,3\% \\
\% m{C_3}{H_7}OH = 100 - 34,3 = 65,7\% \\
b)\\
n{C_6}{H_2}B{r_3}OH = n{C_6}{H_5}OH = 0,1\,mol\\
m = m{C_6}{H_5}B{r_3}OH = 0,1 \times 331 = 33,1g\\
c)\\
nKOH = 0,1 \times 1,6 = 0,16\,mol\\
{C_6}{H_5}OH + KOH \to {C_6}{H_5}OK + {H_2}O\\
n{C_6}{H_5}OK = n{C_6}{H_5}OH = 0,1\,mol\\
nKOH = 0,16 - 0,1 = 0,06\,mol\\
x = 0,1 \times 132 + 0,06 \times 56 = 16,56g
\end{array}\)
