giúppppppppppppp em với ạaaaaaaa

Đáp án:

\(19.\)

\(V_{C_2^{}H_5^{}OH}^{} = 115.75\%  = 86,25\,ml\)

\( \to V_{H_2^{}O}^{} = 115 - 86,25 = 28,75\,lit\)

\( \to m_{H_2^{}O}^{} = V.D = 28,75.1 = 28,75\,gam\)

\(n_{H_2^{}O}^{} = \frac{{28,75}}{{18}} \approx 1,6mol\)

\(m_{C_2^{}H_5^{}OH}^{} = V.D = 0,8.86,25 = 69\,gam\)

\(n_{C_2^{}H_5^{}OH}^{} = \frac{{69}}{{46}} = 1,5\,mol\)

\(2C_2^{}H_5^{}OH + 2Na \to 2C_2^{}H_5^{}ON{\text{a}} + H_2^{}\)

\(\,\,1,5\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,75\)

\(2H_2^{}O + 2Na \to 2NaOH + H_2^{}\)

\(\,\,1,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,8\)

\(V_{H_2^{}}^{} = 22,4.(0,8 + 0,75) = 34,72\,lit\)

\(20.\)

\(C_2^{}H_4^{} + H_2^{}O\xrightarrow{{{t^o},H_2^{}SO_{4_d^{}}^{}}}C_2^{}H_5^{}OH\)

\(C_2^{}H_5^{}OH + O_2^{}\xrightarrow{{xt,{t^o}}}CH_3^{}COOH + H_2^{}O\)

\(CH_3^{}COOH + C_2^{}H_5^{}OH\underset{{{t^o},H_2^{}SO_{4_d^{}}^{}}}{\overset{{{t^o},H_2^{}SO_{4_d^{}}^{}}}{\longleftrightarrow}}CH_3^{}COOC_2^{}H_5^{} + H_2^{}O\)