Em chưa làm được tất cả luôn sao?

 PTHH oxit bazo với nước:

\(Na_2O+H_2O\rightarrow2NaOH\\ K_2O+H_2O\rightarrow2KOH\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Li_2O+H_2O\rightarrow2LiOH\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

Viết 10 PTHH của oxit bazơ tác dụng vs axit

\(NaOH+HCl\rightarrow NaCl+H_2O\\ NaOH+HNO_3\rightarrow NaNO_3+H_2O\\ NaOH+HBr\rightarrow NaBr+H_2O\\ Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\\ Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\\ Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\\ 3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O\\ H_2SO_4+Mg\left(OH\right)_2\rightarrow MgSO_4+2H_2O\\ KOH+HCl\rightarrow KCl+H_2O\\ KOH+HNO_3\rightarrow KNO_3+H_2O\)

phải là 2,24 lít khí `O_2` chứ bạn, nếu không thì rắn A tác dụng với HCl không tạo khí: )

a

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

0,2<--0,1->0,2

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,1->0,2------>0,1-------->0,1

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,2--->0,2------>0,2

b

\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Xét: \(\dfrac{0,3}{2}>\dfrac{0,1}{1}\Rightarrow Mg.dư\)

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

\(V_C=V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c

\(n_{HCl.đã.dùng}=\dfrac{100.29,2\%}{100\%}:36,5=0,8\left(mol\right)\)

\(n_{HCl.dư}=0,8-0,4=0,4\left(mol\right)\)

Dung dịch B gồm: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,1+0,2=0,3\left(mol\right)\\n_{HCl.dư}=0,8-0,4=0,4\left(mol\right)\end{matrix}\right.\)

\(m_{ddB}=m_{Mg.dư}+m_{MgO}+m_{dd.HCl}-m_{H_2}=0,1.24+0,2.40+100-0,1.2=110,2\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0,3.95.100\%}{110,2}=25,86\%\\ C\%_{HCl}=\dfrac{0,4.36,5.100\%}{110,2}=13,25\%\)

Oxit tác dụng được với nước:

Oxit tác dụng được với HCl:

Oxit tác dụng được với CO2:

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ 2Ca + O_2 \xrightarrow{t^o} 2CaO\\ 4Fe_3O_4 + O_2 \xrightarrow{t^o} 6Fe_2O_3\\ Ca + 2H_2O \to Ca(OH)_2 + H_2\\ H_2O + Na_2O \to 2NaOH\)

Kiểm tra lại đề bài 

1.

\(4Na+O_2\underrightarrow{t^0}2Na_2O\)

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)

\(2Ca+O_2\underrightarrow{t^0}2CaO\)

2. 

\(P+3H_2\underrightarrow{t^0}2PH_3\)

\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)

\(Fe_3O_4+4H_2\underrightarrow{t^0}3Fe+4H_2O\)

3. 

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

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

\(K_2O+H_2O\rightarrow2KOH\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(3NO_2+H_2O\rightarrow2HNO_3+NO\)

1) Na,Al,Ca,P

\(4Na + O_2 \xrightarrow{t^o} 2Na_2O\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 2Ca + O_2 \xrightarrow{t^o} 2CaO\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \)

2) CuO,Fe₃O₄

\(CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\)

3) Na,Ca,K₂O,P₂O₅,CaO,NO₂

\(2Na + 2H_2O \to 2NaOH + H_2\\ Ca + 2H_2O \to Ca(OH)_2 + H_2\\ K_2O + H_2O \to 2KOH\\ P_2O_5 + 3H_2O \to 2H_3PO_4\\ CaO + H_2O \to Ca(OH)_2\\ 3NO_2 + H_2O \to 2HNO_3 + NO\)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)

\(0.1.......0.075.....0.05\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(m_{Fe_2O_3}=0.05\cdot160=8\left(g\right)\)

PTHH: \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\)

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

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{Fe_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)

P2O5 + 3H2O => 2H3PO4

H2 + 1/2 O2 => H2O

Na2O + H2O => 2NaOH

K + H2O => KOH + 1/2 H2

Fe2O3 + 3H2 => 2Fe + 3H2O

2K + 1/2 O2 => K2O

2Zn + O2 => 2ZnO

Fe2O3 + 6HCl => 2FeCl3 + 3H2O

Zn + 2HCl => ZnCl2 + H2

Câu 1

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

Câu 2

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\\ Fe_2O_3+H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

NaCl + H2O → NaOH + Cl2 + H2.

refer

TK:

NaCl + H2O → NaOH + Cl2 + H2.

Mik tưởng bằng HOHO chứ