Đáp án:

 6. a) 751,3 g

b) 18,8 kg

7. a) 200 g

b) 5,63%

Giải thích các bước giải:

 6.

a) ${C_2}{H_5}OH + {O_2}\xrightarrow{{}}C{H_3}COOH + {H_2}O$

$\begin{gathered}
  {V_{{C_2}{H_5}OH}} = 10.\dfrac{8}{{100}} = 0,8(l) = 800ml \Rightarrow {m_{{C_2}{H_5}OH}} = D.V = 640\left( g \right) \hfill \\
   \Rightarrow {n_{{C_2}{H_5}OH}} = \dfrac{{640}}{{46}} = 13,91mol \hfill \\
  H = 92\%  \Rightarrow {n_{{C_2}{H_5}OHpu}} = 13,91.90\%  = 12,52mol \hfill \\
   \Rightarrow {n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 12,52mol \hfill \\
   \Rightarrow {m_{C{H_3}COOH}} = 12,52.60 = 751,3\left( g \right) \hfill \\ 
\end{gathered} $

b) ${m_{ddgiam}} = \dfrac{{{m_{C{H_3}COOH}}}}{{4\% }} = \dfrac{{751,3}}{{4\% }} = 18782(g) \approx 18,8kg$

7. 

a) 

$\begin{gathered}
  C{H_3}COOH + NaHC{O_3} \to C{H_3}COONa + C{O_2} + {H_2}O \hfill \\
  {m_{C{H_3}COOH}} = \dfrac{{100.12}}{{100}} = 12(g) \to {n_{C{H_3}COOH}} = 0,2mol \hfill \\
   \Rightarrow {n_{NaHC{O_3}}} = {n_{C{H_3}COOH}} = 0,2mol \Rightarrow {m_{NaHC{O_3}}} = 0,2.84 = 16,8\left( g \right) \hfill \\
   \Rightarrow {m_{ddNaHC{O_3}}} = \dfrac{{16,8}}{{8,4\% }} = 200\left( g \right) \hfill \\ 
\end{gathered} $

b) $\begin{gathered}
  {n_{C{O_2}}} = {n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = 0,2mol \Rightarrow {m_{C{O_2}}} = 0,2.44 = 8,8\left( g \right) \hfill \\
  {m_{C{H_3}COONa}} = 0,2.82 = 16,4(g) \hfill \\ 
\end{gathered} $

Bảo toàn khối lượng: 

$\begin{gathered}
  {m_{ddC{H_3}COOH}} + {m_{ddNaHC{O_3}}} = {m_{ddsau}} + {m_{C{O_2}}} \hfill \\
   \Rightarrow {m_{ddsau}} = {m_{ddC{H_3}COOH}} + {m_{ddNaHC{O_3}}} - {m_{C{O_2}}} \hfill \\
   = 100 + 200 - 8,8 = 291,2\left( g \right) \hfill \\
   \Rightarrow C{\% _{C{H_3}COONa}} = \dfrac{{16,4}}{{291,2}}.100\%  = 5,63\%  \hfill \\ 
\end{gathered} $