a, Ta có: \(n_{NaOH}=\dfrac{30.20\%}{40}=0,15\left(mol\right)\)

PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

__________0,15______0,15 (mol)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

b, \(n_{Na_2CO_3}=0,2.0,5=0,1\left(mol\right)\)

PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,1}{1}\), ta được Na₂CO₃ dư.

Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)

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%

\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,25}{90\%}=\dfrac{5}{18}\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH}=\dfrac{5}{18}.46=\dfrac{115}{9}\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{\dfrac{115}{9}}{0,8}=15,97\left(ml\right)\)

\(\Rightarrow V_{rượu10^o}=\dfrac{15,97}{10\%}=159,7\left(ml\right)\)

\(\left(1\right)C_2H_4+H_2O\xrightarrow[rượu]{men}C_2H_5OH\\ \left(2\right)C_2H_5OH+O_2\xrightarrow[giấm]{men}CH_3COOH+H_2O\\ \left(3\right)\left(4\right)\left(5\right)CH_3COOH+C_2H_5OH⇌_{(H_2SO_4.đặc,t^0)}CH_3COOC_2H_5+H_2O\\ \left(5\right)2CH_3COOH+Mg\xrightarrow[]{}\left(CH_3COO\right)Mg+H_2\)

\(n_{Br_2}=\dfrac{16}{160}=0,1mol\\ C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\\ n_{C_2H_4}=n_{Br_2}=0,1mol\\ V_{C_2H_4}=0,2.22,4=4,48l\\ V_{CH_4}=11,2-4,48=6,72l\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ 2CH_3COOH+Zn\xrightarrow[]{}\left(CH_3COO\right)_2Zn+H_2\\ n_{CH_3COOH}=2n_{H_2}=0,1mol\\ C_{\%CH_3COOH}=\dfrac{0,1.60}{75}\cdot100\%=8\%\)

14.

a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b, \(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,05\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05.22,4}{6,72}.100\%\approx16,67\%\\\%V_{CH_4}\approx83,33\%\end{matrix}\right.\)

15.

\(n_{CO_2}=\dfrac{0,66}{44}=0,015\left(mol\right)=n_C\)

\(n_{H_2O}=\dfrac{0,27}{18}=0,015\left(mol\right)\Rightarrow n_H=0,015.2=0,03\left(mol\right)\)

⇒ m_C + m_H = 0,015.12 + 0,03.1 = 0,21 (g) < m_A

→ A gồm C, H và O.

⇒ m_O = 0,45 - 0,21 = 0,24 (g) 

\(\Rightarrow n_O=\dfrac{0,24}{16}=0,015\left(mol\right)\)

Gọi CTPT của A là C_xH_yO_z.

⇒ x:y:z = 0,015:0,03:0,015 = 1:2:1

→ A có dạng (CH₂O)_n

Mà: M_A = 30.6 = 180 (g/mol)

\(\Rightarrow n=\dfrac{180}{12+1.2+16}=6\)

→ A là C₆H₁₂O₆

bài 14:

a) CH4 + Br2 ------x-----> 

    C2H4 + Br2----------->C2H4Br2

Mol: 0,05 <---0,05 <---------- 0,05

b) nC2H4Br2 = m/M=9,4/188=0,05 (mol)

VC2H4=n . 22,4 = 1,12(mol)

VCH4= 6,72 - 1,12=5,6(mol)

%VC2H4=VC2H4 . 100/6,72 =16,7 %

%VCH4= 100%-16,7%=83,3%

Ta có: \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{0,71}{142}=0,005\left(mol\right)\)

PT: \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

__________0,01_______________0,005______0,005 (mol)

a, \(C_{M_{CH_3COOH}}=\dfrac{0,01}{0,025}=0,4\left(M\right)\)

\(V_{H_2}=0,005.22,4=0,112\left(l\right)\)

b, \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

\(n_{NaOH}=n_{CH_3COOH}=0,01\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,01}{0,75}=\dfrac{1}{75}\left(l\right)=13,33\left(ml\right)\)