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

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

b, \(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\)

Theo PT: \(n_{C_2H_5OH}=n_{CO_2}=n_{CaCO_3}=0,5\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH}=0,5.46=23\left(g\right)\)

\(n_{C_6H_{12}O_6\left(LT\right)}=\dfrac{1}{2}n_{CO_2}=0,25\left(mol\right)\)

Mà: H = 80%

\(\Rightarrow n_{C_6H_{12}O_6\left(TT\right)}=\dfrac{0,25}{80\%}=0,3125\left(mol\right)\)

\(\Rightarrow m_{C_6H_{12}O_6}=0,3125.180=56,25\left(g\right)\)

a, \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

\(CaSO_3+2HCl\rightarrow CaCl_2+SO_2+H_2O\)

\(MgSO_3+2HCl\rightarrow MgCl_2+SO_2+H_2O\)

\(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)

\(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)

b, \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

\(CaSO_3+H_2SO_4\rightarrow CaSO_4+SO_2+H_2O\)

\(MgSO_3+H_2SO_4\rightarrow MgSO_4+SO_2+H_2O\)

\(BaCO_3+H_2SO_4\rightarrow BaSO_4+CO_2+H_2O\)

\(2KHCO_3+H_2SO_4\rightarrow K_2SO_4+2CO_2+2H_2O\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

KCl không pư với NaOH

Ba(NO₃)₂ không pư với NaOH

\(MgSO_4+2NaOH\rightarrow Na_2SO_4+Mg\left(OH\right)_2\)

Na₃PO₄ không pư với NaOH

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

_____0,2______________0,2___0,2 (mol)

a, \(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

\(m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)

b, m _{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

\(\Rightarrow C\%_{ZnSO_4}=\dfrac{32,2}{212,6}.100\%\approx15,15\%\)

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%

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

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

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

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

a, \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

b, Ta có: 24n_Mg + 65n_Zn = 4,245 (1)

Theo PT: \(n_{CH_3COOH}=2n_{Mg}+2n_{Zn}=0,2.1=0,2\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,055\left(mol\right)\\n_{Zn}=0,045\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,055.24=1,32\left(g\right)\\m_{Zn}=0,045.65=2,925\left(g\right)\end{matrix}\right.\)

c, \(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

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

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

\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)

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

__________0,02__________________0,01_____0,01 (mol)

a, \(C_{M_{CH_3COOH}}=\dfrac{0,02}{0,05}=0,4\left(M\right)\)

b, \(V_{H_2}=0,01.22,4=0,224\left(l\right)\)

a, Ta có: \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)

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

________0,2______0,6 (mol)

\(\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)

b, Ta có: V_C2H5OH = 9,2:0,8 = 11,5 (ml)

⇒ \(8=\dfrac{11,5}{V_{rượu8^o}}.100\)

⇒ V _{rượu 8 độ} = 143,75 (ml)

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

Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=0,2\left(mol\right)\)

Mà: H = 80%

\(\Rightarrow n_{CH_3COOH\left(TT\right)}=0,2.80\%=0,16\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=0,16.60=9,6\left(g\right)\)

- Trích mẫu thử.

- Cho từng mẫu thử pư với dd AgNO₃/NH₃

+ Có tủa trắng bạc: glucozo.

PT: \(C_5H_{11}O_5CHO+2AgNO_3+3NH_3+H_2O\underrightarrow{t^o}C_5H_{11}O_5COONH_4+2NH_4NO_3+2Ag\)

+ Không hiện tượng: rượu etylic.

- Dán nhãn.