\(n_{CH_3COOH}=0.5\cdot1=0.5\left(mol\right)\)

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

\(0.5...........................................0.5\)

\(m_{CH_3COOH}=0.5\cdot60=30\left(g\right)\)

\(m_{CH_3COONa}=0.5\cdot82=41\left(g\right)\)

\(n_{Na_2CO_3}=0.2\cdot0.5=0.1\left(mol\right)\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

\(0.2.........................0.1\)

\(\Rightarrow CH_3COOHdư\)

\(n_{CO_2}=n_{Na_2CO_3}=0.1\left(mol\right)\)

\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)

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%

\(m_{CH_3COOH}=24\%.150=36\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{36}{60}=0,6\left(mol\right)\)

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

                0,6                  0,3                  0,6                  0,3 

=> V_CO2 = 0,3.22,4 = 6,72 (l)

\(m_{Na_2CO_3}=0,3.31,8\left(g\right)\)

=> \(m_{ddNa_2CO_3}=\dfrac{31,8}{21,2\%}=150\left(g\right)\)

m_CO2 = 0,3.44 = 13,2 (g)

\(m_{dd}=150+150-13,2=286,8\left(g\right)\)

\(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{24,6}{286,8}=8,58\%\)

Ta có: \(n_{NaOH}=0,1.0,5=0,05\left(mol\right)\)

PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{CH_3COONa}=n_{NaOH}=0,05\left(mol\right)\)

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

b, \(m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)

\(Na_2CO_3(0,05)+2CH_3COOH(0,1)--->2CH_3COONa(0,1)+CO_2(0,05)+H_2O\)

\(m_{Na_2CO_3}=\dfrac{10,6.50}{100}=5,3\left(g\right)\)

\(\Rightarrow n_{Na_2CO_3}=0,05\left(mol\right)\)

Phản ứng vừa đủ:

Theo PTHH: \(nCH_3COOH=0,1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=60.0,1=6\left(g\right)\)

\(b)\)

Theo PTHH: \(n_{CO_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{CO_2}=2,2\left(g\right)\)

\(m_{ddsau}=50+50-2,2=97,8\left(g\right)\)

Dung dich muối tạo thành là CH3COONa

Theo PTHH: \(n_{CH_3COONa}=0,1\left(mol\right)\)

\(\Rightarrow m_{CH_3COONa}=82.0,1=8,2\left(g\right)\)

\(\Rightarrow C\%CH_3COONa=\dfrac{8,2}{97,8}.100\%=8,38\%\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

\(2:3:1:3\left(mol\right)\)

\(0,1:0,15:0,05:0,15\left(mol\right)\)

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

\(a,m_{Al}=n.M=0,1.27=2,7\left(kg\right)\)

\(b,m_{Al_2\left(SO_4\right)_3}=n.M=0,05.342=17,1\left(g\right)\)

PTHH: \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+H_2O+CO_2\uparrow\)

Tính theo sản phẩm

Ta có: \(\left\{{}\begin{matrix}\Sigma n_{CH_3COOH}=\dfrac{160\cdot15\%}{60}=0,4\left(mol\right)\\n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_3COOH\left(dư\right)}=0,2\left(mol\right)=n_{CH_3COONa}\\n_{Na_2CO_3}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na_2CO_3}=0,1\cdot106=10,6\left(g\right)\\m_{CH_3COONa}=0,2\cdot82=16,4\left(g\right)\\m_{CH_3COOH\left(dư\right)}=0,2\cdot60=12\left(g\right)\\m_{CO_2}=0,1\cdot44=4,4\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na_2CO_3}+m_{ddAxit}-m_{CO_2}=166,2\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CH_3COONa}=\dfrac{16,4}{166,2}\cdot100\%\approx9,87\%\\C\%_{CH_3COOH\left(dư\right)}=\dfrac{12}{166,2}\cdot100\%\approx7,22\%\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=0,5\cdot0,3=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối

PTHH: \(CO_2+NaOH\rightarrow NaHCO_3\)

             a_______a__________a      (mol)

            \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

              b_______2b_________2b             (mol)

Ta lập được HPT \(\left\{{}\begin{matrix}a+b=0,1\\a+2b=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,05\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaHCO_3}=0,05\cdot84=4,2\left(g\right)\\m_{Na_2CO_3}=0,05\cdot106=5,3\left(g\right)\end{matrix}\right.\)

CH3COOH+Na->CH3COONa+0,5H2

0,3--------------------------0,3

=>n CH3COOH=0,3 mol

=>m CH3COONa=0,3.82=24,6g