Đáp án+Giải thích các bước giải: 

Bài 1:

2Cu+O2--->2CuO

Mg+2HCl---->MgCl2+H2

2SO2+O2--->2SO3

2P2O5+6H2O---->4H3PO4

2H3PO4+3Ca(OH)2---->Ca3(PO4)2+6H2O

2Al+3H2SO4---->Al2(SO4)3+3H2

2CH4+4O2--->2CO2+4H2O

2Fe(OH)3+3H2SO4---->Fe2(SO4)3+6H2O

2A+2nHNO3--->2A(NO3)n+nH2

Bài 2:

a)PTHH:  3Fe+2O2--->Fe3O4

b) Ta có:nFe=8,4/56= 0,15 mol

Theo PTHH ta có:

nFe3O4=1/3*0,15= 0,05 mol

=>mFe3O4=0,05*232= 11,6 g

c)Theo PTHH ta có:

nO2=2/3*0,15= 0,1 mol

=>VO2=0,1*22,4= 2,24 lít

Bài 3:

a) Zn+2HCl--->ZnCl2+H2

b) Ta có: nZn=19,5/65= 0,3 mol

Theo PTHH ta có:

nH2=nZn= 0,3 mol

=>VH2=0,3*22,4= 6,72 lít

c) Theo PTHH ta có:

nHCl=2*0,3= 0,6 mol

=>mHCl=0,6*36,5= 21,9 g

Bài 4:

PTHH: Mg+2HCl--->MgCl2+H2

a)Ta có: nMg=2,4/24= 0,1 mol

Theo PTHH ta có :

nH2=nMg= 0,1 mol

=>VH2=0,1*22,4= 2,24 lít

b) Theo PTHH ta có:

nHCl=2*0,1= 0,2 mol

=>mHCl=0,2*36,5= 7,3 g

$\text{Đáp án+Giải thích các bước giải:}$

$\text{II Tự luận:}$

$\text{Bài 1:}$

$\text{2Cu+$O_{2}$→2CuO}$

$\text{Mg+2HCl→$MgCl_{2}$+$H_{2}$}$

$\text{$2SO_{2}$+$O_{2}$→$2SO_{3}$}$

$\text{$2P_{2}O_{5}$+$6H_{2}O$→$4H_{3}PO_{4}$}$

$\text{$2H_{3}PO_{4}$+$3Ca(OH)_{2}$→$Ca_{3}(PO_{4})_{2}$+$6H_{2}O$}$

$\text{2Al+$3H_{2}SO_{4}$→$Al_{2}(SO_{4})_{3}$+$3H_{2}$}$

$\text{$CH_{4}$+$2O_{2}$→$CO_{2}$+$2H_{2}O$}$

$\text{$2Fe(OH)_{3}$+$3H_{2}SO_{4}$→$Fe_{2}(SO_{4})_{3}$+$6H_{2}O$}$

$\text{Bài 2:}$

$\text{Câu 1:}$

$\text{A.}$

$\text{PTHH: 3Fe+$2O_{2}$→$Fe_{3}O_{4}$}$

$\text{B.}$

$\text{nFe=$\dfrac{8,4}{56}$= 0,15 mol}$

$\text{Theo PT: $nFe_{3}O_{4}$=$\dfrac{1}{3}$×nFe=$\dfrac{1}{3}$×0,15= 0,05 mol }$

$\text{$mFe_{3}O_{4}$=n×M=0,05×232= 11,6 g}$

$\text{C.}$ 

$\text{Theo PT: $nO_{2}$=$\dfrac{2}{3}$×nFe=$\dfrac{2}{3}$×0,15= 0,1 mol}$

$\text{$VO_{2}$=0,1×22,4= 2,24 lít}$

$\text{Câu 3:}$

$\text{A.}$

$\text{PTHH: Zn+2HCl→$ZnCl_{2}$+$H_{2}$}$

$\text{B.}$

$\text{nZn=$\dfrac{19,5}{65}$= 0,3 mol}$

$\text{Theo PT: $nH_{2}$=nZn= 0,3 mol}$

$\text{$VH_{2}$=n×22,4=0,3×22,4= 6,72 lít}$

$\text{C.}$

$\text{Theo PT: nHCl=2×nZn=2×0,3= 0,6 mol}$

$\text{mHCl=n×M=0,6×36,5= 21,9 g}$

$\text{Câu 14:}$

$\text{A.}$

$\text{PTHH: Mg+2HCl→$MgCl_{2}$+$H_{2}$}$

$\text{B.}$

$\text{nMg=$\dfrac{2,4}{24}$= 0,1 mol}$

$\text{Theo PT: $nH_{2}$=nMg= 0,1 mol}$

$\text{$VH_{2}$=n×22,4=0,1×22,4= 2,24 lít}$

$\text{C.}$

$\text{Theo PT: $nMgCl_{2}$=nMg= 0,1 mol}$

$\text{$mMgCl_{2}$=0,1×95= 9,5 g}$

$\text{D.}$

$\text{Theo PT: nHCl=2×nMg=2×0,1= 0,2 mol}$

$\text{mHCl=n×M=0,2×36,5= 7,3 g}$

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