1)
a) Gọi số mol \(C_2H_5OH\) và \(C_3H_7OH\) lần lượt là \(x;y\)
\( \to 46x + 60y = 10,6{\text{ gam}}\)
Phản ứng xảy ra:
\({C_2}{H_5}OH + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 3{H_2}O\)
\(2{C_3}{H_7}OH + 9{O_2}\xrightarrow{{{t^o}}}6C{O_2} + 8{H_2}O\)
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
Ta có:
\({n_{CaC{O_3}}} = \frac{{50}}{{100}} = 0,5{\text{ mol = }}{{\text{n}}_{C{O_2}}} = 2x + 3y\)
Giải được: \(x=y=0,1\)
\( \to {m_{{C_2}{H_5}OH}} = 0,1.46 = 4,6{\text{ gam}}\)
\({m_{{C_3}{H_7}OH}} = 0,1.60 = 6{\text{ gam}}\)
\({n_{{H_2}O}} = 0,1.3 + 0,1.4 = 0,7{\text{ mol}}\)
\( \to m = {m_{C{O_2}}} + {m_{{H_2}O}} = 0,5.44 + 0,7.18 = 34,6{\text{ gam}}\)
b)Dẫn hỗn hợp \(X\) qua \(CuO\)
\(C{H_3}C{H_2}OH + CuO\xrightarrow{{{t^o}}}C{H_3}CHO + Cu + {H_2}O\)
\(C{H_3}C{H_2}C{H_2}OH + CuO\xrightarrow{{{t^o}}}C{H_3}C{H_2}CHO + Cu + {H_2}O\)
2)
a)Phản ứng xảy ra:
\(2{C_2}{H_5}OH + 2Na\xrightarrow{{}}2{C_2}{H_5}ONa + {H_2}\)
\(2{C_6}{H_5}OH + 2Na\xrightarrow{{}}2{C_6}{H_5}ONa + {H_2}\)
b)\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \to {{\text{n}}_{{C_2}{H_5}OH}} + {n_{{C_6}{H_5}OH}} = 2{n_{{H_2}}} = 0,3{\text{ mol}}\)
Cho hỗn hợp tác dụng với \(Br_2\)
\({C_6}{H_5}OH + 3B{r_2}\xrightarrow{{}}{C_6}{H_2}B{r_3}OH + 3HBr\)
Ta có:
\({n_{{C_6}{H_2}B{r_3}OH}} = \frac{{19,86}}{{12.6 + 2 + 80.3 + 17}} = 0,06{\text{ mol = }}{{\text{n}}_{{C_6}{H_5}OH}}\)
\( \to {n_{{C_2}{H_5}OH}} = 0,24{\text{ mol}}\)
\( \to {m_{{C_2}{H_5}OH}} = 0,24.46 = 11,04{\text{ gam}}\)
\({m_{{C_6}{H_5}OH}} = 0,06.(12.6 + 5 + 17) = 5,64{\text{ gam}}\)
\( \to \% {m_{{C_2}{H_5}OH}} = \frac{{11,04}}{{11,04 + 5,64}} = 66,2\% \to \% {m_{{C_6}{H_5}OH}} = 33,8\% \)
