a. MnO2 +4HCl→Cl2+2H2O+MnCl2

2Fe+3Cl2→2FeCl3

FeCl3+3NaOH → Fe(OH)3 +3NaCl

Fe(OH)3+3NaCl→ FeCl3 +3NaOH

FeCl3+3AgNO3→3AgCl+Fe(NO3)3

2AgCl→Cl2+2Ag

b. 2KMnO4 +16HCl→ 5Cl2+8H2O+2KCl+2MnCl2

Cl2+H2→2HCl

CuO+2HCl→CuCl2+H2O

CuCl2+Ba(OH)2→ BaCl2+Cu(OH)2

BaCl2+H2SO4→ BaSO4.+2HCl

c.2 NaCl+H2SO4→ 2HCl +Na2SO4

2HCl→ Cl2 +H2

3Cl2+2Fe→ 2FeCl3

FeCl3+3NaOH →3NaCl +Fe(OH)3

2NaCl+2H2O→2NaOH+Cl2+H2

NaOH+HCl→NaCl+H2O

2NaCl→Cl2+2Na

Cl2+Ca →CaCl2

CaCl2+2AgNO3→2AgCl +Ca(NO3)2

2AgCl→2Ag+Cl2

D

Cặp chất 2 và 4 không xảy ra phản ứng.

Cặp chất 1 và 3 xảy ra phản ứng với nhau.

PTHH: C a C l 2 + N a 2 C O 3 → C a C O 3 ↓ + 2 N a C l

N a O H + H C l → N a C l + H 2 O

⇒ Chọn D.

2NaCl + 2H₂O \(\underrightarrow{đpddcmn}\)2NaOH + Cl₂ + H₂

Cl₂ + H₂ \(\rightarrow\)2HCl

Fe₂O₃ + 6HCl \(\rightarrow\)2FeCl₃ + 3H₂

2FeCl₃ + Cu \(\rightarrow\)CuCl₂ + 2FeCl₂

số mol NaOH là:\(n_{NaOH}=\frac{10}{23+16+1}=0,25\left(mol\right)\)

PTHH\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(m_{FeCl_3}=n.M=\frac{0.25}{3}\cdot\left(56+35,5\cdot3\right)\approx13,54\left(g\right)\)

\(m_{Fe\left(OH\right)_3}=n.M=\frac{0.25}{3}\cdot\left(56+\left(16+1\right)\cdot3\right)\approx8,91\left(g\right)\)

\(m_{NaCl}=n.M=0.25\cdot\left(23+35.5\right)=14.625\left(g\right)\)

n(NaOH) = m(NaOH)/M(NaOH) = 50/40 =1.25

n(HCl) = m(HCl)/M(HCl) = 36.5/36.5 =1.25

Gọi x là số mol của NaCl

PTHH: NaOH + HCl \(\rightarrow\) NaCl + H2O

Theo phương trình: 1 mol 1 mol 1 mol

Theo đề: 1.25 mol 1 mol x mol

Ta có tỉ lệ : 1.25/1 > 1/1 \(\Rightarrow\) NaOH dư

HCl hết

n(NaCl) = x =1*1/1 = 1 mol

m(NaCl) = n(NaCl) * M(NaCl) = 1* 58.5 =58.5 (g)

PTHH. NaOH + HCl -> NaCl + H2O

Theo bài:
nNaOH/bđ = \(\dfrac{50}{40}=1,25\) (mol)

nHCl/ bđ =\(\dfrac{36,5}{36,5}=1\left(mol\right)\)

Theo pthh:

nHCl/ pt = 1(mol)
nNaOH/ pt = 1(mol)

Ta có tỉ lệ:

\(\dfrac{nNaOH_{bd}}{nNaOH_{pt}}=\dfrac{1,25}{1}=1,25>\dfrac{nHCl_{pt}}{nHCl_{pt}}=\dfrac{1}{1}=1\)

Sau pư NaOH còn dư, HCl tgpu hết
=> nNaCl = nHCl = 1 (mol)
=>mNaCl = nNaCl . MNaCl = 1 . 58,5 = 58,5 (g)

\(n_{NaOH}=\dfrac{50}{40}=1,25\left(mol\right)\)
\(n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Theo PTHH ta có:
\(\dfrac{1,25}{1}>\dfrac{1}{1}\)
\(\Rightarrow NaOH\) dư, HCl phản ứng hết nên ta tính theo \(n_{HCl}\)
Theo PT ta có: \(n_{NaCl}=1\left(mol\right)\)
\(\Rightarrow m_{NaCl}=1.58,5=58,5\left(g\right)\)

Câu 1:

1.

H2 + Cl2 => 2HCl

H2 + HCl => NaCl + H2

2.

2NaCl => 2Na + Cl2

Cl2 + H2 => 2HCl

Mg + HCl => MgCl2 + H2

3.

2NaCl => 2Na + Cl2

3Cl2 + 2Fe => 2FeCl3

Câu 2:

1.

Ta có:

\(\left\{{}\begin{matrix}n_{SO2}=\frac{12,8}{64}=0,2\left(mol\right)\\n_{NaOH}=\frac{1.250}{1000}=0,25\left(mol\right)\end{matrix}\right.\)

PTHH:

\(SO_2+NaOH\rightarrow NaHSO_4\)

x________x _________x____

\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)

y _______2y________ y ________________

Tỉ lệ : \(1\le\frac{n_{NaOH}}{n_{SO2}}\le2\Leftrightarrow1\le\frac{0,25}{0,2}\le2\)

Giải hệ PT:

\(\left\{{}\begin{matrix}x+y=0,2\\x+2y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,05\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaHSO3}=0,15.104=15,6\left(g\right)\\m_{Na2SO3}=0,5.126=6,3\left(g\right)\end{matrix}\right.\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(2Na+Cl_2\rightarrow2NaCl\)

\(2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl\)

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

\(MgCl_2+2AgNO_3\rightarrow2AgCl+Mg\left(NO_3\right)_2\)

\(2AgCl\rightarrow2Ag+Cl_2\)