a) \(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄(đ),t^o--> CH₃COOC₂H₅ + H₂O

                  0,5------->0,5------------------------>0,5

=> m_C2H5OH = 0,5.46 = 23 (g)

b) m_CH3COOC2H5 = 0,5.88 = 44 (g)

a, \(n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\)

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

Theo PT: \(n_{Mg}=n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow m=m_{Mg}=0,1.24=2,4\left(g\right)\)

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

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

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

\(\Rightarrow m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)

\(\Rightarrow V_{ddC_2H_5OH}=\dfrac{9,2}{0,8}=11,5\left(ml\right)\)

\(a)n_{CH_3COOH} = 0,2.2 = 0,4(mol)\\ Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{Mg} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ m_{Mg} = 0,2.24 = 4,8(gam)\\ b)\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ n_{CH_3COOH\ pư} = n_{este} = \dfrac{24,64}{88} = 0,28(mol)\\ H = \dfrac{0,28}{0,4}.100\% = 70\%\)

\(n_{CH_3COOH}=\dfrac{300\cdot5\%}{60}=0.25\left(mol\right)\)

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

\(0.25........................................................0.125\)

\(V_{H_2}=0.125\cdot22.4=2.8\left(l\right)\)

\(n_{C_2H_5OH}=0.1\cdot2=0.2\left(mol\right)\)

\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\left(ĐK:H_2SO_{4\left(đ\right)},t^0\right)\)

\(0.2......................0.2.....................0.2\)

\(\Rightarrow CH_3COOHdư\)

\(m_{CH_3COOC_2H_5}=0.2\cdot88=17.6\left(g\right)\)

a) n CH3COOH = 300.5%/60 = 0,25(mol) 

Zn + 2CH3COOH $\to$ (CH3COO)2Zn + H2

Theo PTHH :

n H2 = 1/2 n CH3COOH = 0,25/2 = 0,125(mol)

V H2 = 0,125.22,4 = 2,8(lít)

b) n C2H5OH = 0,1.2 = 0,2(mol)

\(CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\)

Ta thấy :

n CH3COOH = 0,25 > n C2H5OH = 0,2  => CH3COOH dư

n CH3COOC2H5 = n C2H5OH = 0,2 mol

=> m CH3COOC2H5 = 0,2.88 = 17,6 gam

a) nCH3COOH= 0,4(mol)

PTHH: CH3COOH + NaOH -> CH3COONa + H2O

0,4____________0,4(mol)

=> mNaOH=0,4. 40=16(g)

b) nCH3COOH= 1(mol)

nC2H5OH= 100/46= 50/23(mol)

Vì : 1/1< 50/23 :1

=> C2H5OH dư, CH3COOH hết, tính theo nCH3COOH.

PTHH: CH3COOH + C2H5OH \(⇌\) CH3COOC2H5 + H2O (đk: H⁺ , nhiệt độ)

Ta có: nCH3COOC2H5(thực tế)= 0,625(mol)

Mà theo LT: nCH3COOC2H5(LT)= nCH3COOH=1(mol)

=>H= (0,625/1).100=62,5%

a) \(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

                 0,1-------->0,2------------->0,2

=> \(V_{dd.CH_3COOH}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

b) \(m_{CH_3COONa}=0,2.82=16,4\left(g\right)\)