7)
Các phản ứng xảy ra:
\(C{H_4} + C{l_2}\xrightarrow{{as}}C{H_3}Cl + HCl\)
\(C{H_3}Cl + NaOH\xrightarrow{{}}C{H_3}OH + NaCl\)
\(C{H_3}OH + CuO\xrightarrow{{}}HCHO + Cu + {H_2}O\)
\(2HCHO + {O_2}\xrightarrow{{{t^o}}}2HCOOH\)
\(C{H_3}OH + CO\xrightarrow{{{t^o},xt}}C{H_3}COOH\)
8)
Gọi số mol \(CH_3COOH;HCOOH\) lần lượt là \(x;y\)
\( \to 60x + 46y = 16,6{\text{ gam}}\)
Cho hỗn hợp tác dụng với \(NaOH\)
\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)
\(HCOOH + NaOH\xrightarrow{{}}HCOONa + {H_2}O\)
Ta có:
\({n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = x;{n_{HCOONa}} = {n_{HCOOH}} = y\)
\( \to 82x + 68y = 23,2\)
Giải được: \(x=0,2;y=0,1\)
\( \to {m_{C{H_3}COOH}} = 0,2.60 = 12{\text{ gam}}\)
\( \to \% {m_{C{H_3}COOH}} = \frac{{12}}{{16,6}} = 72,29\% \to \% {m_{HCOOH}} = 27,71\% \)
