CH3COOH+NaOH->CH3COONa+H2O

0,4---------------0,4

n CH3COOH=0,4 mol

=>CM NaOH=\(\dfrac{0,4}{0,1}=4M\)

$a\big)$

$n_{CH_3COOH}=\dfrac{100}{1000}.1=0,1(mol)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo PT: $n_{NaOH}=n_{CH_3COOH}=0,1(mol)$

$\to C\%_{NaOH}=\dfrac{0,1.40}{50}.100\%=80\%$

$b\big)$

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

$2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O$

Theo PT: $\begin{cases} n_{CO_2}=n_{Na_2CO_3}=0,1(mol)\\ n_{CH_3COONa}=2n_{Na_2CO_3}=0,2(mol) \end{cases}$

$\to C\%_{CH_3COONa}=\dfrac{0,2.82}{60+10,6-0,1.44}.100\%\approx 24,77\%$

a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PT: \(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{KOH}=0,15\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

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) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

PTHH: 2CH₃COOH + Mg --> (CH₃COO)₂Mg + H₂

                  0,1<----------------------0,05------->0,05

=> V_H2 = 0,05.22,4 = 1,12 (l)

b) \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,1}{0,025}=4M\)

\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)

PTHH: CH₃COOH + NaOH ---> CH₃COONa + H₂O

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

\(m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=\dfrac{8}{8,4\%}=\dfrac{2000}{21}\left(g\right)\\ m_{ddCH_3COONa}=\dfrac{2000}{21}+100=\dfrac{4100}{21}\left(g\right)\\ m_{CH_3COONa}=0,2.82=16,4\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{16,4}{\dfrac{4100}{21}}.100\%=8,4\%\)

`=>` Gợi ý:

`CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O`

`m<sub>CH3COOH</sub> = 100x12/100 = 12` (g)

`==> n<sub>CH3COOH</sub> = m/M = 12/60 = 0.2` (mol)

Theo pt: `=> n<sub>NaHCO3</sub> = 0.2` (mol)

`==> m<sub>NaHCO3</sub> = n.M = 0.2x84 =16.8` (g)

`==> m<sub>dd NaHCO3</sub> = 16.8x100/8.4 = 200` (g)

Ta có: `n<sub>CH3COONa</sub> = 0.2` (mol)

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, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

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

Theo PT: \(n_{CH_3COOH}=2n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

b, \(n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4\left(g\right)\)