Đáp án:
\( m = 11,7{\text{ gam}}\)
Giải thích các bước giải:
\({C_6}{H_5}OH + 3B{r_2}\xrightarrow{{}}{C_6}{H_2}B{r_3}OH + 3HBr\)
Ta có:
\({m_{B{r_2}}} = 480.10\% = 48{\text{ gam}}\)
\( \to {n_{B{r_2}}} = \frac{{48}}{{80.2}} = 0,3{\text{ mol}}\)
\( \to {n_{{C_6}{H_5}OH}} = \frac{1}{3}{n_{B{r_2}}} = 0,1{\text{ mol}} \to {{\text{n}}_{{C_2}{H_5}OH}} = 0,05{\text{ mol}}\)
\( \to m = {m_{{C_6}{H_5}OH}} + {m_{{C_2}{H_5}OH}} = 0,1.94 + 0,05.46 = 11,7{\text{ gam}}\)