Chương I. Các loại hợp chất vô cơ

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\)

PTHH: \(Fe+S-^{t^o}\rightarrow FeS\)

Theo đề: 0,2...0,3

Lập tỉ lệ \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\)=> Fe hết. S dư

=> Sau phản ứng hỗn hợp gồm S dư, FeS

=> \(n_{S\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\)

S + H2SO4 → SO2 + H2O

2FeS + 10H2SO4 → Fe2(SO4)3 + 9SO2 + 10H2O

\(\Rightarrow m_{H_2SO_4}=\left(0,1+\dfrac{0,2.10}{2}\right).98=107,8\left(g\right)\)

\(V_{SO_2}=\left(0,1+\dfrac{0,2.9}{2}\right).22,4=22,4\left(l\right)\)

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{FeS}=0,1\left(mol\right)\)

3Ba(OH)2 + Fe2(SO4)3 → 2Fe(OH)3↓ + 3BaSO4↓ (*)

Theo PT (*) : \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,05\left(mol\right)\)

Theo PT (**): \(n_{Fe_2O_3}=2n_{Fe\left(OH\right)_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_2O_3}=0,1.160=16\left(g\right)\)

\(3KOH+AlCl_3\rightarrow3KCl+Al\left(OH\right)_3\)

\(NaAlO_2+HCl+H_2O\rightarrow Al\left(OH\right)_3+NaCl\)

\(NaAlO_2+CO_2+2H_2O\rightarrow Al\left(OH\right)_3+NaHCO_3\)

\(2AlCl_3+3Na_2CO_3+3H_2O\rightarrow2Al\left(OH\right)_3+6NaCl+3CO_2\)

\(AlF_3+3H_2O\rightarrow Al\left(OH\right)_3+3HF\)

\(Al_4C_3+12H_2O\rightarrow4Al\left(OH\right)_3+3CH_4\)

$AlCl_3+NaOH\rightarrow Al(OH)_3+NaCl$

$NaAlO_2+CO_2+H_2O\rightarrow Al(OH)_3+NaHCO_3$

$AlCl_4+Na_2CO_3+H_2O\rightarrow Al(OH)_3+CO_2+NaCl$

$Al_2(CO_3)_3+H_2O\rightarrow Al(OH)_3+CO_2$

$AlN+H_2O\rightarrow Al(OH)_3+NH_3$

$Al_4C_3+H_2O\rightarrow Al(OH)_3+CH_4$

$AlCl_3 + 3KOH \to Al(OH)_3 + 3KCl$
$NaAlO_2 + HCl + H_2O \to Al(OH)_3 + NaCl$

$2AlCl_3 + 3Na_2CO_3 + 3H_2O \to 2Al(OH)_3 + 3CO_2 + 6NaCl$

$NaAlO_2 + NH_4Cl + H_2O \to NH_3 + Al(OH)_3 + NaCl$

$2Al + 6H_2O \to 2Al(OH)_3 + 3H_2$

$NaAlO_2 + CO_2 + 2H_2O \to Al(OH)_3 + NaHCO_3$

\(n_{H_2SO_4}=0.3\cdot1.5=0.45\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(0.9..............0.45\)

\(m_{NaOH}=0.9\cdot40=36\left(g\right)\)

\(n_{KOH}=0.9\left(mol\right)\)

\(m_{KOH}=0.9\cdot56=50.4\left(g\right)\)

\(m_{dd_{KOH}}=\dfrac{50.4}{5.6\%}=900\left(g\right)\)

\(V_{ddKOH}=\dfrac{900}{1.045}=861.2\left(ml\right)\)

Giup minh voi

a)

$Cu + 2H_2SO_{4_{đặc}} \to CuSO_4 + SO_2 + 2H_2O$

$n_{SO_2} = n_{Cu} = \dfrac{6,4}{64} = 0,1(mol)$
$V = 0,1.22,4 = 2,24(lít)$

b)

$n_{CuSO_4} = n_{Cu} = 0,1(mol)$
$m_{CuSO_4} = 0,1.160 = 16(gam)$

c)

$n_{H_2SO_4} = 2n_{Cu} = 0,2(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,2.98}{90\%} = 21,78(gam)$

\(n_{Cu}=\dfrac{6.4}{64}=0.1\left(mol\right)\)

\(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

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

\(V_{SO_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{CuSO_4}=0.1\cdot160=16\left(g\right)\)

\(m_{dd_{CuSO_4}}=\dfrac{0.2\cdot98\cdot100}{90}=21.78\left(g\right)\)

n_CO2=13,44/22,4=0,6(mol)

m_CO2=0,6.44=26,4(g)

PTHH:CO₂+Ca(OH)₂-->CaCO₃

dùng ĐLBTKL ta có:

m_CO2+m_Ca(OH)2=m_CaCO3

m_Ca(OH)2=74,8-26,4=48,4(g)

-->n_Ca(OH)2=48,4/74=0,65(mol)

-->V_Ca(OH)2=0,65/2,4=0,27(lít)

nếu sai rất mong mọi người giúp đỡ

+) Những p/ứ 2, 5, 6, 7, 8 là những p/ứ oxi hóa - khử vì có sự thay đổi số oxi hóa của những chất tham gia p/ứ

+) Trong p/ứ 2: \(\overset{0}{Fe}\rightarrow\overset{+2}{Fe}\) và \(\overset{+1}{H}\rightarrow\overset{0}{H}\)

+) Trong p/ứ 5 và 7: \(\overset{+2}{Fe}\rightarrow\overset{+3}{Fe}\) và \(\overset{+6}{S}\rightarrow\overset{+4}{S}\)

+) Trong p/ứ 6: \(\overset{0}{Fe}\rightarrow\overset{+3}{Fe}\) và \(\overset{+6}{S}\rightarrow\overset{+4}{S}\)

+) Trong p/ứ 8: \(\overset{+2}{Fe}\overset{-2}{S}\rightarrow\overset{+3}{Fe}+\overset{+6}{S}\) và \(\overset{+6}{S}\rightarrow\overset{+4}{S}\)

2) 

\(Fe\rightarrow Fe^{+2}+2e\)

\(S^{+6}+2e\rightarrow S^{+4}\)

5) 

\(Fe^{+2}\rightarrow Fe^{+3}+1e\)

\(S^{+6}+2e\rightarrow S^{+4}\)

6) 

\(Fe\rightarrow Fe^{+3}+3e\)

\(S^{+6}+2e\rightarrow S^{+4}\)

7) 

\(Fe^{+2}\rightarrow Fe^{+3}+1e\)

\(S^{+6}+2e\rightarrow S^{+4}\)

8) 

\(Fe^{+2}\rightarrow Fe^{+3}+1e\)

\(S^{+6}+2e\rightarrow S^{+4}\)

4 PHẢN ỨNG CUỐI NHÉ BẠN

5) Fe²⁺ \(\rightarrow\) Fe³⁺  + 1e

6) Fe⁰  \(\rightarrow\) Fe³⁺ + 3e

7) Fe²⁺ \(\rightarrow\) Fe³⁺  + 1e

8) Fe²⁺ \(\rightarrow\) Fe³⁺  + 1e

- Phản ứng 1 ,3

Phản ứng (1) : 

\(3FeO+2Al\rightarrow Al_2O_3+3Fe\)

\(Fe^{+2}+2e\rightarrow Fe\)

\(Al\rightarrow Al^{+3}+3e\)

Phản ứng (3) : 

\(4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\)

\(Fe^{+2}\rightarrow Fe^{+3}+1e\)

\(O_2+4e\rightarrow2O^{2-}\)

\(SO_3:\)\(S^{+6}\)

\(K_2SO_4:K^{+1},S^{+6},O^{-2}\)

\(P_2O_5:P^{+5},O^{-2}\)

\(H_3PO_4:H^{+1},P^{+5},O^{-2}\)

\(FeO:Fe^{+2},O^{-2}\)

\(Fe_3O_4:Fe^{+\dfrac{8}{3}},O^{-2}\)

\(H_2S:H^{+1},S^{-2}\)

\(CH_4:C^{-4},H^{+1}\)

\(HNO_3:H^{+1},N^{+5},O^{-2}\)

\(NO:N^{+2},O^{-2}\)

\(NO_2:N^{+4},O^{-2}\)

\(N_2O:N^{+1},O^{-2}\)

\(FeS:Fe^{+2},S^{-2}\)

\(Fe\left(NO_3\right)_3:Fe^{+3}\)

\(Al_2\left(SO_4\right)_3:Al^{+3}\)

\(CaCO_3:Ca^{+2}\)

SO3 : S+6

K2SO4 : K+1; S+6

P2O5 : P+5

H3PO4 : P+5

FeO : Fe+2

Fe3O4 : Fe+8/3

H2S : S-2

CH4 : C-4

HNO3 : N+5

NO : N+2

NO2 : N+4

N2O : N+1

FeS : Fe+2; S-2

Fe(NO3)3 : Fe+3 , N+5

Al2(SO4)3 : Al+3; S+6

CaCO3 : Ca+2; C+4

O có số oxi hóa là -2

H có số oxi hóa là +1

a)

$MCO_3 \xrightarrow{t^o} MO + CO_2$
$CO_2 + Ba(OH)_2 \to BaCO_3 + H_2O$
$CO_2 + Ba(OH)_2 \to Ba(HCO_3)_2$

b)

$n_{Ba(OH)_2} = 0,15(mol) ; n_{BaCO_3} = \dfrac{19,7}{197} = 0,1(mol)$

$n_{Ba(HCO_3)_2} = n_{Ba(OH)_2} - n_{BaCO_3} = 0,05(mol)$
$n_{CO_2} = n_{BaCO_3} + 2n_{Ba(HCO_3)_2} = 0,2(mol)$

$n_{MCO_3} = n_{CO_2} = 0,2(mol)$

$m_A = m_{MCO_3} - m_{CO_2} = 20 - 0,2.44 = 11,2(gam)$
c)
$M_{MCO_3} = M + 60 = \dfrac{20}{0,2} = 100$

$\Rightarrow M = 40(Canxi)$

Vậy CTHH : $CaCO_3$

a, Ta có: $m_{CaSO_4}=0,2(g);m_{H_2O}=100(g)$

$\Rightarrow \%C_{CaSO_4}=0,2\%$

Mặt khác $V_{ddCaSO_4}=100,2(ml)\Rightarrow C_{M/CaSO_4}=0,015M$

b, Ta có: $n_{CaCl_2}=0,006(mol);n_{Na_2SO_4}=0,002(mol)$

$\Rightarrow n_{CaSO_4}=0,002(mol)\Rightarrow m=0,272(g)$

Giả sử 200ml dung dịch là $H_2O$ $\Rightarrow m_{dd}=200(g)$

So sánh với độ tan của $CaSO_4$ thì không có kết tủa xuất hiện