a, Ta có: \(n_{CH_3COOH}=\dfrac{100.15\%}{60}=0,25\left(mol\right)\)
PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)
___________0,25_____0,125_________0,25____0,125 (mol)
\(\Rightarrow m_{ddK_2CO_3}=\dfrac{0,125.138}{13,8\%}=125\left(g\right)\)
b, m dd sau pư = 100 + 125 - 0,125.44 = 219,5 (g)
\(\Rightarrow C\%_{CH_3COOK}=\dfrac{0,25.98}{219,5}.100\%\approx11,61\%\)
c, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
\(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,25\left(mol\right)\)
Mà: H = 90%
\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,25}{90\%}=\dfrac{5}{18}\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH}=\dfrac{5}{18}.46=\dfrac{115}{9}\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{\dfrac{115}{9}}{0,8}=15,97\left(ml\right)\)
\(\Rightarrow V_{rượu10^o}=\dfrac{15,97}{10\%}=159,7\left(ml\right)\)