\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

\(n_{CO_2}=\frac{0,6}{22,4}=\frac{3}{112}mol\)

\(n_{KOH}=2n_{CO_2}=\frac{3}{56}mol\)

\(CM_{KOH}=\frac{3}{56}:\frac{100}{1000}\approx0,356M\)

hình như sai đề....

\(\ast C_2H_6O\)

– CTCT 1:

– CTCT 2:

\(\ast C_2H_5NO_2\)

– CTCT 1:

– CTCT 2:

– CTCT 3:

\(\ast C_2H_6O\):

 – CTCT 1 : \(CH_3\text{ – }CH_2\text{ – }OH\) (Rượu etylic)

– CTCT 2 : \(CH_3-O-CH_3\) (Đimetyl ete)

\(\ast C_2H_5NO_2\) 

– CTCT 1 : \(CH_3-CH_2-NO_2\) (Nitroetan)

– CTCT 2 : \(CH_3-CH_2-O-N=O\)(Etyl nitrit)

– CTCT 3 : \(NH_2-CH_2-COOH\)(Glyxin)

- Axit:

\(H_2SO_3\) - Axit sunfurơ

\(H_2SO_4\) - Axit sunfuric

\(H_3BO_3\) - Axit boric

\(HNO_3\) - Axit nitric

\(HClO\) - Axit hipoclorơ

\(H_4TiO_4\) - Axit titanic

- Bazơ:

\(Fe\left(OH\right)_2\) - Sắt(II) hiđroxit

\(NaOH\) - Natri hiđroxit

\(KOH\) - Kali hiđroxit

\(Cu\left(OH\right)_2\) - Đồng(II) hiđroxit

\(LiOH\) - Liti hiđroxit

- Muối:

\(NaCl\) - Natri clorua

\(KCl\) - Kali clorua

\(NaI\) - Natri iotua

\(LiNO_3\) - Liti nitrat

\(CaCO_3\) - Canxi cacbonat

\(NH_4NO_3\) - Amoni nitrat

a)

Zn+2HCl→ZnCl2+H2

CuO+H2→Cu+H2O

b)

Chất oxi hóa : CuO

Chất khử : H2

nZn=3,25/65=0,05 moln

Theo PT ta có nH2=nZn=0,05 mol

nCuO=680=0,075 mol

CuO+H2→Cu+H2O

Theo PT ta có tỉ lệ 0,075/1>0,05/1

→CuO dư 

→nCu=nH2=0,05 mol

→mCu=0,05.64=3,2 g

c)

CuO dư

nCuOphản ứng =nH2=0,05 mol

nCuO dư =0,075−0,05=0,025 mol

mCuO dư =0,025.80=2 g

\(CaHPO_4+NaHSO_4\rightarrow CaSO_4+NaH_2PO_4\)

Bài 1 :

Gọi 

\(n_{Fe} = a(mol) ; n_{Zn} = b(mol)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Zn + 2HCl \to ZnCl_2 + H_2\\ \)

Ta có : 

\(\hept{\begin{cases}n_{H_2}=a+b=\frac{3,36}{22,4}=0,15\left(mol\right)\\m_{muoi}=127a+136b=19,5\left(gam\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=0,1\\b=0,05\end{cases}}\)\(\Rightarrow\hept{\begin{cases}m_{Fe}=0,1.56=5,6\left(gam\right)\\m_{Zn}=0,05.65=3,25\left(gam\right)\end{cases}}\)

Bài 2 : 

\(\hept{\begin{cases}n_{BaCO_3}=a\left(mol\right)\\n_{BaSO_3}=b\left(mol\right)\end{cases}}\)

\(BaCO_3 + 2HCl \to BaCl_2 + CO_2 + H_2O\\ BaSO_3 + 2HCl \to BaCl_2 + SO_2 + H_2O\)

Ta có : 

\(\hept{\begin{cases}m_{hh}=197a+217b=20,5\left(gam\right)\\n_{khí}=n_{CO_2}+n_{SO_2}=a+b=\frac{2,24}{22,4}=0,1\left(mol\right)\end{cases}}\)

Suy ra:  a = 0,06 ; b = 0,04

\(\%m_{BaCO_3} = \dfrac{0,06.197}{20,5}.100\% =57,66\%\\ \%m_{BaSO_3} = 100\%- 57,66\%=42,34\%\)

\(n_{FeCl_2}=0,25mol\)

Đặt \(\hept{\begin{cases}x=n_{Fe}\\y=n_{FeO}\\z=n_{FeCO_3}\end{cases}}\)

Có \(\frac{2x+44z}{x+z}=30\rightarrow-28x+14z=0\)

BT Fe \(x+y+z=0,25\)

Có hệ phương trình

\(\hept{\begin{cases}56x+72y+116z=21,6\\x+y+z=0,25\\-28x+14z=0\end{cases}}\)

\(\rightarrow\hept{\begin{cases}x=0,05\\y=0,1\\z=0,1\end{cases}}\)

\(\%m_{Fe}=\frac{0,05.56}{21,6}.100\%=12,9\%\)

\(\%m_{FeO}=\frac{0,1.72}{21,6}.100\%=33,3\%\)

\(\rightarrow\%m_{FeCO_3}=100\%-12,9\%-33,3\%=53,8\%\)