\(n C a C O 3 = 10 100 = 0 , 1 ( m o l ) P T H H : C a C O 3 + 2 H C l → C a C l 2 + C O 2 ↑ + H 2 O ⇒ n H C l = 2 n C a C O 3 = 0 , 2 ( m o l ) ⇒ m H C l = 0 , 2 ⋅ 36 , 5 = 7 , 3 ( g )\)

Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

\(PTHH:CaCO_3+2HCl--->CaCl_2+CO_2+H_2O\)

Theo PT: \(n_{HCl}=2.n_{CaCO_3}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)

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

\(2Na+Cl_2\rightarrow2NaCl\)

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

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

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

\(2AgCl\rightarrow2Ag+Cl_2\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Pt : \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3|\)

         2         3           2

        0,2     0,3

\(n_{Cl2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\)

\(V_{Cl2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

 Chúc bạn học tốt

A: H₂

B: HCl

C: FeCl₂

D: FeCl₃

E: NaOH

F: Fe(OH)₃

G: Fe₂O₃

H: H₂O

PTHH:

(1): Cl₂ + H₂ =(nhiệt)=> 2HCl

(2): 2HCl + Fe ===> FeCl₂ + H₂

(3): 2FeCl₂ + Cl₂ ===> 2FeCl₃

(4): FeCl₃ + 2NaOH ===> Fe(OH)₃\(\downarrow\) + 2NaCl

(5): 2Fe(OH)₃ =(nhiệt)=> Fe₂O₃ + 3H₂O

(6): Fe₂O₃ + 3H₂ =(nhiệt)=> 2Fe + 3H₂O

\(Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ \Rightarrow m_{H_2}=2.0,1=0,2\left(g\right);m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

Fe + 2HCl → FeCl₂ + H₂

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,15\times56=8,4\left(g\right)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT ta có: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
Khối lượng Fe đã phản ứng là:
\(m_{Fe}=0,15.27=8,4\left(g\right)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT ta có: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

a. PTHH: Zn + 2HCl → ZnCl2 + H2

nZn = 13 : 65 = 0,2 (mol)

b.Theo PT: nHCl = 2 nZn = 2. 0,2 = 0,4 mol

Vậy mHCl = 0,4 . 36,5 = 14,6(g)

c. Theo PTHH: nH2 = nZn = 0,2 mol

VH2 (đktc) = 0,2 . 22,4 = 4,48 (l)

a,PTHH: Zn+2HCl--->ZNCl2+H2

b,nZn=0,2mol

=>2nZn = nHCl = 0,4 mol

=>mHCl= 0,4.36,5= 14,6 (g)

c,nH2 = 0,2 (nH2 = nZn)

=>VH2 = 4,48l

1

a) 2Al + 6HCl ---> 2AlCl₃ +3H₂

b) 2AgNO₃ + BaCl₂ ---> 2AgCl + Ba(NO₃)₂

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(2AgNO_3+BaCl_2\rightarrow2AgCl+Ba\left(NO_3\right)_2\)

a,

Ca: kim loại

CaO: oxit bazo

Ca(OH)\(_2\) : bazo

CaCl\(_2\) : Muối

HCl: axit

Cl\(_2\) : khí clo

b, \(2Ca+O_2\rightarrow2CaO\)

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

\(Ca\left(OH\right)_2+CuCl_2\rightarrow Cu\left(OH\right)_2+CaCl_2\)

\(CaCl_2+H_2\rightarrow Ca+2HCl\)

\(4HCl+MnO_2\rightarrow MnCl_2+Cl_2+2H_2O\)

a, Ca: kim loại

CaO: oxit bazơ

Ca(OH)₂ : dd Bazơ

CaCl₂ :Muối

HCl: Axit

Cl₂ : Phi kim

b, 2Ca +O₂ --^to--> 2CaO

CaO + H₂O ----> Ca(OH)₂

Ca(OH)₂ + MgCl₂ ----> CaCl₂ +Mg(OH)₂

CaCl₂ +H₂SO₄ -----> CaSO₄ + 2HCl

6HCl + 2Al -------> 2AlCl₃ +3H₂

xin lỗi, mình đính chính phương trình cuối :

2KnO4 +16 HCl ---^_to-> 2KCl + 5Cl₂ + 8H₂O