Cho 11,2 lit CO₂ vào 500ml dd NaOH 25% (d = 1,3g/ml). Tính nồng độ mol/lit của dd muối tạo thành.

2KClO3 ---> 2KCl + 3O2

4P +5O2 ---> 2P2O5

P2O5 + 3H2O --->2H3PO4

2Fe(OH)3 ---> Fe2O3 +3H2O

2Fe(OH)3 +3H2SO4 ---> Fe2(SO4)3 +6H2O

tớ làm được câu 5 rồi hằng ạ!

a, Cac chat co the dieu che hidro la

Zn , Al , Cu , H2O tac dung voi dd HCl va H2SO4

Phuong trinh hoa hoc

Zn + HCl \(\rightarrow\) ZnCl2 + H2

Zn + H2SO4 \(\rightarrow\) ZnSO4 + H2

2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2

2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2

Cu + H2SO4 \(\rightarrow\) CuSO4 + H2

Cu + 2HCl \(\rightarrow\) CuCl2 + H2

2H2O\(\underrightarrow{dienphan}\) 2H2 + O2

b, Cac chat co the dung de dieu che O2 la

H2O , KMnO4 , KClO3

KMnO4 \(\underrightarrow{t0}\) K2MnO4 + MnO2 + O2

KClO3 \(\underrightarrow{to}KCl+O2\)

H2O \(\underrightarrow{dienphan}\) H2 + O2

a) \(3Fe+2O_2-t^o->Fe_3O_4\)

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

Theo pthh : \(n_{Fe_3O_4}=\frac{1}{3}n_{Fe_3O_4}=\frac{0,1}{3}\left(mol\right)\)

=> \(m_{Fe_3O_4}=232\cdot\frac{0,1}{3}\approx7,73\left(g\right)\)

c) Theo pthh : \(n_{O2\left(pứ\right)}=\frac{2}{3}n_{Fe}=\frac{0,2}{3}\left(mol\right)\)

=> \(n_{O2\left(can.dung\right)}=\frac{0,2}{3}\div100\cdot120=0,08\left(mol\right)\)

=> \(V_{O2\left(can.dung\right)}=0,08\cdot22,4=1,792\left(l\right)\)

a, Ta có n_O2 = \(\dfrac{3,2}{32}\) = 0,1 ( mol )

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

0,2................................................0,1

=> m_{KMnO4 cần dùng để điều chế 3,2 g oxi} = 158 . 0,2 = 31,6 ( gam )

2KClO₃ → 2KCl + 3O₂

\(\dfrac{1}{15}\)............................0,1

=> m_{KCl cần dùng để điều chế 3,2 gam oxi} = 122,5 . \(\dfrac{1}{15}\) = 8,17 ( gam )

b,

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

0,1................................................0,05

=> m_O2 = 0,05 . 32 = 1,6 ( gam )

2KClO₃ → 2KCl + 3O₂

0,1............................0,15

=> m_O2 = 32 . 0,15 = 4,8 ( gam )

Ta có n_KMnO4 = \(\dfrac{50}{158}\) = \(\dfrac{25}{79}\) ( mol )

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

\(\dfrac{25}{79}\)................................................\(\dfrac{25}{158}\)

=> m_O2 = \(\dfrac{25}{158}\) . 32 \(\approx\) 5,06 ( gam )

n_KClO3 = \(\dfrac{50}{122,5}\) = 0,408 ( mol )

2KClO₃ → 2KCl + 3O₂

0,408......................0,612

=> m_O2 = 0,612 . 32 = 19,584 ( gam )

a) nFe=0,45mol

PTHH: 2Fe+O2=>2FeO

              0,45->0,225

=> VO2 cần dùng =0,225.22,4=5,04 lít

b)2KClO3=>2KCl+3O2

      0,15<---------------0,225

=> mKClO3=0,15.122,5=18,375g

A.

Số mol của Fe: n=\(\frac{m}{M}\) =\(\frac{25,2}{56}\) = 0.45 (mol)

2Fe + O₂ --t⁰-> 2FeO

Theo PT 2 : 1 : 2

Theo bài ra 0.45 : 0.225 : 0.45 (mol)

Thể tích Oxi tham gia phảm ứng: V = n . 22,4 = 5.04 ( lít )

B.

Ta có: 2KClO₃ -t⁰-> 2KCl + 3O₂

Theo PT 2 : 2 : 3

Theo bài ra 0,15 : 0,15 : 0,225 (mol)

Khối lượng KClO₃ : m = n.M = 0.15 . 122,5 = 18,375 (g)

1. \(2KCl_3\underrightarrow{^{xt}}2KCl+3O_2\)

2. \(2H_2+O_2\underrightarrow{^{t^o}}2H_2O\)

3.\(SO_3+H_2O\rightarrow H_2SO_4\)

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

5.\(PbO+H_2\underrightarrow{^{t^o}}Pb+H_2O\)

6.\(Pb+O_2\underrightarrow{^{t^p}}PbO\)

7.\(2KMnO_4\underrightarrow{^{t^o}}K_2MnO_4+MnO_2+O_2\)

8.\(3Fe+2O_2\underrightarrow{^{t^o}}Fe_3O_4\)

9.\(Na_2O+H_2O\rightarrow2NaOH\)

10.\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

a) \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\uparrow\)

\(n_{Ca}=\frac{8}{40}=0,2mol\)

Theo phương trình \(n_{H_2}=n_{Ca}=0,2mol\)

\(\rightarrow V_{H_2}=0,2.22,4=4,48l\)

b) Theo phương trình \(n_{Ca\left(OH\right)_2}=n_{Ca}=0,2mol\)

\(\rightarrow m_{Ca\left(OH\right)_2}=0,2.\left(40+17.2\right)=14,8g\)

\(m_{ddsaupu}=m_{Ca}+m_{H_2O}-m_{H_2}\)

\(\rightarrow m_{ddsaupu}=8+200-0,2.2=207,6g\)

\(\rightarrow C\%_{ddCa\left(OH\right)_2}=\frac{14,8.100}{207,6}=7,13\%\)

a, 4Al+3O2-> 2Al2O3

b, 3H2O +P2O5 -> 2H3PO4

c, 2K+ 2H2O -> 2KOH + H2

d, K2O + H2O -> 2KOH

e, KMnO4 -> K2MnO4 + MnO4 + O2

g, CuO+ H2-> Cu+H2O

h, FexOy+H2-> xFe +yH2O