\(m_{CH_3COOH}=150.30\%=45\left(g\right)\\ n_{CH_3COOH}=\dfrac{45}{60}=0,75\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH \(\xrightarrow[t^o]{H_2SO_4đặc}\) CH₃COOC₂H₅ + H₂O

             0,75 ----------> 0,75 ------------------> 0,75

\(V_{C_2H_5OH}=\dfrac{46.0,75}{0,8}=43,125\left(ml\right)\\ m_{CH_3COOC_2H_5}=0,75.45\%.88=29,7\left(g\right)\)

B5:

\(n_{CuSO_4}=0,2.0,1=0,02\left(mol\right)\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ Cu\left(OH\right)_2\rightarrow\left(t^o\right)CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,02\left(mol\right)\\ m_D=m_{CuO}=0,02.80=1,6\left(g\right)\\ m_{Na_2SO_4}=142.0,02=2,84\left(g\right)\)

• Trần Hữu Tuyển62GP
• Hồ Hữu Phước35GP
• Ngô Thanh Sang16GP
• Azue6GP
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• nguyen khanh duy

Cl manhj hon CO3 mak banj ???

Ta có Cu không tác dụng được H₂SO₄ loãng

=> Chất rắn D là Cu

Mg + H₂SO₄ → MgSO₄ + H₂

x...........x.................x..............x

Fe + H₂SO₄ → FeSO₄ + H₂

y.........y................y............y

MgSO₄ + 2KOH → Mg(OH)₂ + K₂SO₄

x................2x.................x.................x

FeSO₄ + 2KOH → Fe(OH)₂ + K₂SO₄

y................2y..............y..................y

4Fe(OH)₂ + O₂ + 2H₂O → 4Fe(OH)₃

y................................................y

Mg(OH)₂ + O₂ ↛

2Fe(OH)₃ → Fe₂O₃ + 3H₂O

y......................y/2..........1,5y

n_Fe(OH)3 = y = \(\dfrac{24}{107}\) ( mol )

n_CuO = \(\dfrac{5}{80}\) = 0,0625 ( mol )

Cu + \(\dfrac{1}{2}\)O₂ → CuO

0,0625...........0,0625

=> m_Cu = 64 . 0,0625 = 4 ( gam )

=> %m_Cu = \(\dfrac{4}{20}\) . 100 = 20 %

=> %m_Fe = \(\dfrac{\dfrac{24}{107}.56}{20}\) . 100 = 62,8 %

=> %m_Mg = 100 - 20 - 62,8 = 17,2 %

em cam on anh

a)

\(Cu + 2H_2SO_4 đ \rightarrow^{t^o} CuSO_4 + SO_2 + 2H_2O\)

\(CuSO_4 + BaCl_2 \rightarrow CuCl_2 + BaSO_4\)

\(CuCl_2 + 2NaOH \rightarrow Cu(OH)_2 + 2NaCl\)

\(Cu(OH)_2 \rightarrow^{t^o} CuO + H_2O\)

\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)

\(CuCl_2 + Fe \rightarrow Cu + FeCl_2\)

b)

\(2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2\)

\(AlCl_3 + 3NaOH_{vừa đủ/thiếu} \rightarrow Al(OH)_3 + 3NaCl\)

\(2Al(OH)_3 \rightarrow^{t^o} Al_2O_3 + 3H_2O\)

\(2Al_2O_3 \rightarrow^{đpnc} 4Al + 3O_2\)

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

\(Al_2(SO_4)_3 + 6NaOH_{vừa đủ/ thiếu} \rightarrow 2Al(OH)_3 + 3Na_2SO_4\)

\(Al(OH)_3 + 3HCl \rightarrow AlCl_3 + 3H_2O\)

c)

\(Fe + H_2SO_4 \rightarrow FeSO_4 + H_2\)

\(FeSO_4 + BaCl_2 \rightarrow BaSO_4 + FeCl_2\)

\(FeCl_2 + 2NaOH \rightarrow Fe(OH)_2 + 2NaCl\)

\(Fe(OH)_2 \rightarrow^{t^o} FeO + H_2O\)

\(FeO + H_{2_{dư}} \rightarrow^{t^o} Fe + H_2O\)

\(Fe + 2HCl \rightarrow FeCl_2 + H_2\)

\(Mg + FeCl_2 \rightarrow MgCl_2 + Fe\)

\(3Fe + 2O_2 \rightarrow^{t^o} Fe_3O_4\)

Bài 1:

c) (1) S+ O2 -> SO2

(2) SO2 +  2 NaOH -> Na2SO3 + H2O

(3) SO2 + 1/2 O2 \(⇌\left(to,xt\right)\) SO3

(4) SO3 + H2O -> H2SO4

(5) H2SO4 + 2 NaOH -> Na2SO4 +2 H2O

(6) Na2SO4 + BaCl2 -> BaSO4 + 2 NaCl

(7) SO3 ra chất gì mũi tên xuống?

(8) SO2 + H2O \(⇌\) H2SO3

(9) H2SO3+ Na2SO4 -> H2SO4 + Na2SO3

(10) Na2SO3+ H2SO4 -> Na2SO4 + SO2 + H2O

Bài 1: 

a) (1) CaCO3 + Ba(OH)2 -> BaCO3 + Ca(OH)2

(2) CaCO3 -to-> CaO + CO2

(3) CaCO3 ra CaCO3?

(4) CaO + 2 HCl -> CaCl2 + H2O