Hóa học

a, Ta có: \(n_{CH_3COOH}=\dfrac{100.15\%}{60}=0,25\left(mol\right)\)

PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)

___________0,25_____0,125_________0,25____0,125 (mol)

\(\Rightarrow m_{ddK_2CO_3}=\dfrac{0,125.138}{13,8\%}=125\left(g\right)\)

b, m _{dd sau pư} = 100 + 125 - 0,125.44 = 219,5 (g)

\(\Rightarrow C\%_{CH_3COOK}=\dfrac{0,25.98}{219,5}.100\%\approx11,61\%\)

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

\(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,25\left(mol\right)\)

Mà: H = 90%

\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,25}{90\%}=\dfrac{5}{18}\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH}=\dfrac{5}{18}.46=\dfrac{115}{9}\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{\dfrac{115}{9}}{0,8}=15,97\left(ml\right)\)

\(\Rightarrow V_{rượu10^o}=\dfrac{15,97}{10\%}=159,7\left(ml\right)\)

\(\left(1\right)C_2H_4+H_2O\xrightarrow[rượu]{men}C_2H_5OH\\ \left(2\right)C_2H_5OH+O_2\xrightarrow[giấm]{men}CH_3COOH+H_2O\\ \left(3\right)\left(4\right)\left(5\right)CH_3COOH+C_2H_5OH⇌_{(H_2SO_4.đặc,t^0)}CH_3COOC_2H_5+H_2O\\ \left(5\right)2CH_3COOH+Mg\xrightarrow[]{}\left(CH_3COO\right)Mg+H_2\)

\(a.Fe_3O_4+8HCl\xrightarrow[]{}FeCl_2+2FeCl_3+4H_2O\)     

\(2NaOH+FeCl_2\xrightarrow[]{}Fe\left(OH\right)_2+2NaCl\\ 3NaOH+FeCl_3\xrightarrow[]{}Fe\left(OH\right)_3+3NaCl\\ NaOH+HCl\xrightarrow[]{}NaCl+H_2O\\ 2Fe\left(OH\right)_3\xrightarrow[]{t^0}Fe_2O_3+3H_2O\\ 4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^0}2Fe_2O_3+4H_2O\)

A gồm \(FeCl_2,FeCl_3,HCl.dư\)

B gồm \(NaCl\)

C gồm \(Fe\left(OH\right)_2,Fe\left(OH\right)_3\)

\(b.BTNT\left(Fe\right):3n_{Fe_3O_4}=2n_{Fe_2O_3}\\ \Leftrightarrow3.0,1=2n_{Fe_2O_3}\\ \Rightarrow n_{Fe_2O_3}=0,15mol\\ m=m_{Fe_2O_3}=0,15.160=24g\)

\(n_{Br_2}=\dfrac{16}{160}=0,1mol\\ C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\\ n_{C_2H_4}=n_{Br_2}=0,1mol\\ V_{C_2H_4}=0,2.22,4=4,48l\\ V_{CH_4}=11,2-4,48=6,72l\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ 2CH_3COOH+Zn\xrightarrow[]{}\left(CH_3COO\right)_2Zn+H_2\\ n_{CH_3COOH}=2n_{H_2}=0,1mol\\ C_{\%CH_3COOH}=\dfrac{0,1.60}{75}\cdot100\%=8\%\)

- AlCl₃: Alumini clorua
- Na₂SO₄: Natri sunfat
- Ba(HCO₃)₂: Bari cacbonat
- FeS: Sắt sulfua
- CuSO₄: Đồng sulfat
- KNO₃: Kali nitrat
- FeSO₄: Sắt sulfat

AlCl₃ alumminium chloride

Na₂SO₄ Sodium sulfate

Ba(HCO₃)₂ Barium hydrogencarbonate

FeS Iron (II) sulfide

CuSO₄ Copper (II) sulfate

KNO₃ Potassium nitrate

FeSO₄ Iron (II) sulfate 

đây nhé

\(n_{MnO_2}=\dfrac{69,6}{87}=0,8\left(mol\right)\)

\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+2Cl_2+2H_2O\)

0,8------------------------------>1,6

Chất oxi hóa là \(MnO_2\)

\(Mn^{+4}+2e\rightarrow Mn^{+2}\)

Chất khử là HCl

\(Cl^{-1}\rightarrow Cl^0+2e\)

\(n_{KOH}=0,5.4=2\left(mol\right)\)

\(3Cl_2+2KOH\rightarrow KCl+KClO+H_2O\)

0,8--->\(\dfrac{8}{15}\)------->\(\dfrac{4}{15}\)---->\(\dfrac{4}{15}\)

\(CM_{KOH}=\dfrac{2-\dfrac{8}{15}}{0,5}=2,93M\)

\(CM_{KCl}=\dfrac{\dfrac{4}{15}}{0,5}=0,5M\\ CM_{KClO}=\dfrac{\dfrac{4}{15}}{0,5}=0,5M\)