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 + b,

MgCl → MgCl₂ - Magie clorua

KSO₃ → K₂SO₃ - kali sunfit

CaHCO₃ → Ca(HCO₃)₂ - canxi hidrocacbonat

AlS → Al₂S₃ - Nhôm sunfua

a, \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\)

b, Không pư.

c, \(Fe+CuCl_2\rightarrow FeCl_2+Cu\)

d, \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_4\)

e, \(Ca\left(OH\right)_2+Na_2CO_3\rightarrow2NaOH+CaCO_3\)

f, Không pư

g, \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

h, \(2NH_4Cl+Ba\left(OH\right)_2\rightarrow BaCl_2+2NH_3+2H_2O\)

1. \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

2. \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

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

4. \(CH\equiv CH+2Br_2\rightarrow CHBr_2-CHBr_2\)

5. \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

6. \(CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

7. \(2CH_3COOH+2Na\rightarrow2CH_3COONa+H_2\)

8. Giống PT 4

Câu 5:

Ta có: \(n_{RO}=\dfrac{6.10^{22}}{6.10^{23}}=0,1\left(mol\right)\)

\(\Rightarrow M_{RO}=\dfrac{5,6}{0,1}=56\left(g/mol\right)\)

⇒ M_R + 16 = 56

⇒ M_R = 40 (g/mol)

→ R là Ca.

Ta có: n_C2H5OH + n_C6H5OH = 2n_H2 = 0,15.2 (1)

n_C6H5OH = n_C6H2Br3OH = 0,1 (mol) (2)

Từ (1) và (2) ⇒ n_C2H5OH = 0,2 (mol)

\(\Rightarrow\%m_{C_2H_5OH}=\dfrac{0,2.46}{0,2.46+0,1.94}.100\%\approx49,46\%\)

- 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.

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%

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

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

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

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

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

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