Ví dụ : Từ câu (a) => (g)

a) \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

b) \(BaCl_2+NaOH\) ----//---->

c) \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\)

\(Ag^++Cl^-\rightarrow AgCl\)

d) \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(Fe^{2+}+2OH^-\rightarrow Fe\left(OH\right)_2\)

e) \(Na_2S+2HCl\rightarrow2NaCl+H_2S\)

\(2H^++S^{2-}\rightarrow H_2S\)

f) \(Na_2CO_3+2HNO_3\rightarrow2NaNO_3+H_2O+CO_2\)

\(2H^++CO_3^{2-}\rightarrow CO_2+H_2O\)

g) \(CuS+HCl\)----//---->

Em cảm ơn chị!!!

1/ Gọi x là hóa trị của nguyên tố cần tìm. Áp dụng quy tắc hóa trị:

a) Fe2O3 : x.2=II.3 => x=III

CuO : x.1=II.1 => x=II

N2O3 : x.2=II.3 => x=III

SO3: x.1=II.3 => x= VI

b) NH3 : x.1=I.3 => x=III

C2H2 : C hóa trị IV, H hóa trị I ( do trong hợp chất hữu cơ, hóa trị của C luôn là IV)

HBr : I.1=x.1 => x=I

H2S: I.2=x.1 => x=II

c)K2S: x.2=II.1 => x=I

MgS : x.1=II.1 => x=II

Cr2S3 : x.2=II.3 => x=III

CS2: x.1=II.2=> x=IV

d) KCl: x.1=I.1=> x=I

HCl: x.1= 1.I => x=I

BaCl2 : x.1=I.2 => x=II

AlCl3 : x.1=I.3 => x=III

e) ZnCO3 : x.2=II.1 => x=II

BaSO4 : x.1=II.1 => x=II

Li2CO3 : x.2=II.1 => x=I

Cr2(SO4)3 : x.2=II.3 => x=III

f) NaOH : x.1=1.I => x=I

Zn(OH)2 : x.1=I.2 => x=II

AgNO3 : x.1=I.1 => x=I

Al(NO3)3 : x.1=I.3 => x=III

2.a) HCl : H(I), Cl(I)

H2S: H(I), S(II)

NH3 : N(III), H(I)

H2O : H(I), O(II)

CH4: C(IV), H(I)

b) NO: N(II), O(II)

N2O: N(I), O(II)

NO2: N(IV), O(II)

N2O5: N(V), O(II)

HNO3 : H(I), NO3 (I)

Ca(NO3)2 : Ca(II), NO3 (I)

NaNO3: Na(I), NO3 (I)

Al(NO3)3: Al (III),NO3 (I)

c) CaO: Ca(II), O(II)

K2O: K(I), O(II)

MgO : Mg(II), O(II)

Na2O: Na(I), O(II)

Al2O3: Al(III), O(II)

d) SO2: S(IV) ,O(II)

SO3: S(VI), O(II)

Na2S: Na(I), S(II)

FeS: Fe(II), S(II)

Al2S3: Al(III), S(II)

H2SO4: H(I), SO4(II)

CuSO4: Cu(II), SO4(II)

Al2(SO4)3: Al(III), SO4(II)

e) P2O5: P(V), O(II)

H3PO4: H(I), PO4(III)

Na3PO4: Na(I), PO4(III)

Ca3(PO4)2: Ca(II), PO4(III)

\(n_{C_2H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Theo PT: \(n_{Br_2}=2n_{C_2H_2}=0,2\left(mol\right)\Rightarrow C_{M_{Br_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

→ Đáp án: C

CaC2+2H2O->C2H2+Ca(OH)2

C2H2+H2-to>C2H4

C2H2+2H2-to>C2H6

C2H4+Br2->C2H4Br2

n-(CH2=CH2-)--to>-(CH2-CH2)-_n

a)

$CH_3COOH + KOH \to CH_3COOK + H_2O$

n CH3COOH = n KOH = 0,1(mol)

$CH_3COOH + 2O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$

$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
$CO_2 +C a(OH)_2 \to CaCO_3 + H_2O$

2n C2H5OH + 2n CH3COOH = n CO2 = n CaCO3 = 60/100 = 0,6(mol)

=> n C2H5OH = (0,6 - 0,1.2)/2 = 0,2(mol)

=> n O2 = 2n CH3COOH + 3n C2H5OH = 0,1.2 + 0,2.3 = 0,8(mol)

=> V O2 = 0,8.22,4 = 17,92 lít

b) m = 0,1.60 + 0,2.46 = 15,2 gam

%m CH3COOH = 0,1.60/15,2  .100% = 39,47%
%m C2H5OH = 100%- 39,47% = 60,53%

Cảm ơn bạn nhiều ạ!😊

33A

34B

35C

36D

37D

38A

39B

40.Natri