\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)

a, Theo PT: \(n_{CH_3COOH}=2n_{Fe}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{10\%}=120\left(g\right)\)

\(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 5,6 + 120 - 0,1.2 = 125,4 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{0,1.174}{125,4}.100\%\approx13,88\%\)

\(a) n_{K_2CO_3} = \dfrac{5,52}{138}=0,04(mol)\\ K_2CO_3 + 2CH_3COOH \to 2CH_3COOK + CO_2 + H_2O\\ n_{CH_3COOH} = 2n_{K_2CO_3} = 0,08(mol)\\ m_{dd\ CH_3COOH} = \dfrac{0,08.60}{12\%} = 40(gam)\\ b) CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\\ n_{CaCO_3} = n_{CO_2} = n_{K_2CO_3} = 0,04(mol)\\ m_{CaCO_3} = 0,04.100 = 4(gam)\)

Câu 4:

a, \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

Theo PT: \(n_{CH_3COOH}=2n_{H_2}=0,5\left(mol\right)\Rightarrow V_{CH_3COOH}=\dfrac{0,5}{2,5}=0,2\left(l\right)\)

b, \(n_{\left(CH_3COO\right)_2Zn}=n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{\left(CH_3COO\right)_2Zn}=0,25.183=45,75\left(g\right)\)

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%

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

=> m_CH3COOH = 0,2.60 = 12 (g)

\(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,4}=0,5M\)

b) \(n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH₃COOH

\(n_{CH_3COOH\left(pư\right)}=\dfrac{0,2.80}{100}=0,16\left(mol\right)\)

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

                 0,16------------------------------------->0,16

=> \(m_{CH_3COOC_2H_5}=0,16.88=14,08\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2......0.4..........0.2........0.2\)

\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)

\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)

\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)

a) \(n_{CH_3COOH}=0,1.0,3=0,03\left(mol\right)\)

PTHH: CH₃COOH + NaOH --> CH₃COONa + H₂O

                  0,03---->0,03--------->0,03

=> \(V_{dd.NaOH}=\dfrac{0,03}{1,5}=0,02\left(l\right)\)

b) m_CH3COONa = 0,03.82 = 2,46 (g)

c) \(C_{M\left(CH_3COONa\right)}=\dfrac{0,03}{0,1+0,02}=0,25M\)

Phần 2: n NaOH = n axit + n este = 0,1 mol Phần 3: n CO2 = n axit = 0,06 mol
Do 2 phần này có khối lượng như nhau nên => mỗi phần có neste = 0,04 mol
=> tổng cả 2 phần này có 0,12 mol axit và 0,08 mol este
Phần 1 giả sử có x mol ancol ; y mol este và 1,5y mol axit
( do tỉ lệ mol các chất trong hỗn hợp như nhau giữa các phần)
=> n ancol trong 2 phần còn lại là 0,08x/y mol => ta có:
Ở phần 1: 0,08x/y + 1,5y = 2nH2 = 0,04 mol
mX = 46(x+ 0,08x/y) + 60(0,12 + 1,5y ) + 88(y + 0,08)
Giải hệ ta được y = 0,02 mol ; x = 0,1 mol
=> m ancol (P1)= 0,46g =>B

PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)

a+b+c) Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KCl}=0,5\left(mol\right)=n_{HCl}\\n_{K_2SO_3}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_3}=0,25\cdot158=39,5\left(g\right)=a\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{ddHCl}=\dfrac{0,5\cdot36,5}{10,95\%}\approx166,67\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{SO_2}=0,25\cdot64=16\left(g\right)\)

\(\Rightarrow m_{dd}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=190,17\left(g\right)\) \(\Rightarrow C\%_{KCl}=\dfrac{37,25}{190,17}\cdot100\%\approx19,59\%\)

d) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Theo PTHH: \(n_{NaOH}=n_{HCl}=0,5\left(mol\right)\) \(\Rightarrow V_{NaOH}=\dfrac{0,5}{0,5}=1\left(l\right)\)

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

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)

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

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