\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

Theo PT: \(n_{O_2}=2n_{CH_4}=0,2\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)

a, \(2CH_3COOH+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2H_2O\)

b, \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

c, \(C_2H_4+H_2O\underrightarrow{t^o,xt}C_2H_5OH\)

d, \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (đk: t^o, H₂SO₄ đặc)

a, \(Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

b, \(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

c, \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

d, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

e, \(CH_4+Cl_2\underrightarrow{as}CH_3Cl+HCl\)

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

a, \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (đk: t^o, H₂SO₄ đặc)

Tên gọi: etyl axetat

b, \(n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,1.88=8,8\left(g\right)\)

\(\Rightarrow H=\dfrac{5,5}{8,8}.100\%=62,5\%\)

a, \(C_6H_{12}O_6\underrightarrow{^{t^o,xt}}2C_2H_5OH+2CO_2\)

Ta có: \(n_{C_6H_{12}O_6}=\dfrac{45}{180}=0,25\left(mol\right)\)

Theo PT: \(n_{C_2H_5OH\left(LT\right)}=2n_{C_6H_{12}O_6}=0,5\left(mol\right)\)

Mà: H = 90%

⇒ n_{C2H5OH (TT)} = 0,5.90% = 0,45 (mol)

\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,45.46=20,7\left(g\right)\)

b, \(V_{C_2H_5OH}=\dfrac{20,7}{0,8}=25,875\left(ml\right)\)

⇒ V _{rượu 20^o} = 25,875:20.100 = 129,375 (ml)

c, \(C_2H_5OH+O_2\underrightarrow{^{xt}}CH_3COOH+H_2O\)

______0,45______________0,45 (mol)

\(\Rightarrow m_{ddCH_3COOH3\%}=\dfrac{0,45.60}{3\%}=900\left(g\right)\)

a, \(2CH_3COOH+Ca\rightarrow\left(CH_3COO\right)_2Ca+H_2\)

\(n_{\left(CH_3COO\right)_2Ca}=\dfrac{3,95}{158}=0,025\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{\left(CH_3COO\right)_2Ca}=0,05\left(mol\right)\)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,05}{0,25}=0,2\left(M\right)\)

b, \(n_{H_2}=n_{\left(CH_3COO\right)_2Ca}=0,025\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,025.22,4=0,56\left(l\right)\)

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

\(n_{NaOH}=n_{CH_3COOH}=0,05\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,05}{0,5}=0,1\left(l\right)=100\left(ml\right)\)

a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(C_6H_5OH+Na\rightarrow C_6H_5ONa+\dfrac{1}{2}H_2\)

\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

Theo PT: \(2n_{H_2}=n_{C_6H_5OH}+n_{C_2H_5OH}=0,4\left(mol\right)\left(1\right)\)

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

PT: \(C_6H_5OH+NaOH\rightarrow C_6H_5ONa+H_2O\)

Theo PT: n_C6H5OH = n_NaOH = 0,2 (mol) (2)

Từ (1) và (2) ⇒ n_C2H5OH = 0,2 (mol)

⇒ m = 0,2.46 + 0,2.94 = 28 (g)

b, \(\%m_{C_2H_5OH}=\dfrac{0,2.46}{28}.100\%\approx32,86\%\)

\(\%m_{C_6H_5OH}\approx67,14\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PT: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

\(C_6H_5OH+Na\rightarrow C_6H_5ONa+\dfrac{1}{2}H_2\)

Theo PT: \(n_{C_2H_5OH}+n_{C_6H_5OH}=2n_{H_2}=0,3\left(mol\right)\left(1\right)\)

PT: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr\)

Có: \(n_{C_6H_2Br_3OH}=\dfrac{19,86}{331}=0,06\left(mol\right)\)

Theo PT: \(n_{C_6H_5OH}=n_{C_6H_2Br_2OH}=0,06\left(mol\right)\) (2)

Từ (1) và (2) ⇒ n_C2H5OH = 0,24 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,24.46}{0,24.46+0,06.94}.100\%\approx66,2\%\\\%m_{C_6H_5OH}\approx33,8\%\end{matrix}\right.\)

a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

b, Ta có: \(n_{C_2H_6O}=\dfrac{9,2}{46}=0,2\left(mol\right)\)

Theo PT: \(n_{CO_2}=2n_{C_2H_6O}=0,4\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,4.22,4=8,96\left(l\right)\)

Ta có: n_H2 = 0,25 (mol)

\(CH_3OH+K\rightarrow CH_3OK+\dfrac{1}{2}H_2\)

\(C_6H_5OH+K\rightarrow C_6H_5OK+\dfrac{1}{2}H_2\)

⇒ n_CH3OH + n_C6H5OH = 2n_H2 = 0,5 (1)

\(n_{KOH}=\dfrac{200.8,4\%}{56}=0,3\left(mol\right)=n_{C_6H_5OH}\) (2)

\(C_6H_5OH+KOH\rightarrow C_6H_5OK+H_2O\)

Từ (1) và (2) ⇒ n_CH3OH = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3OH}=\dfrac{0,2.32}{0,2.32+0,3.94}.100\%\approx18,5\%\\\%m_{C_6H_5OH}\approx81,5\%\end{matrix}\right.\)