Zn+H2SO4->ZnSO4+H2

a,nH2=2,24/22,4=0,1(mol)

=>nZn=0,1(mol)

=>mZn pư=0,1*65=6,5(g)

b.nH2SO4=0,1(mol)

VH2SO4=0,1/2,5=0,04(l)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH:

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,1       0,1             0,1        0,1

\(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

\(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

\(a,PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\ \Rightarrow n_{Zn}=0,1\left(mol\right)\\ \Rightarrow m_{Zn}=0,1\cdot65=6,5\left(g\right)\\ b,n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,1\cdot98=9,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{9,8\cdot100\%}{10\%}=98\left(g\right)\)

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

Zn + H2SO4 -> ZnSO4 + H2

Câu 1 : 

\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(0.02......................0.02\)

\(m_{dd}=1.6+100=101.6\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{101.6}\cdot100\%=3.15\%\)

Câu 2 : 

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2......0.4..........0.2........0.2\)

\(m_{dd_{HCl}}=\dfrac{0.4\cdot36.5\cdot100}{8}=182.5\left(g\right)\)

\(m_{dd}=13+182.5-0.2\cdot2=195.1\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)

a, PTHH: Zn+H2SO4--->ZnSO4+H2 (1)

nH2= \(\dfrac{6,72}{22,4}=0,3\) mol

Theo pt(1): nZn=nH2= 0,3 mol

=>mZn= 0,3.65= 19,5 (g)

b, Đổi 100ml= 0,1 lit

Theopt(1): nH2SO4=nH2= 0,3mol

=> CM_H2SO4= \(\dfrac{0,3}{0,1}=3M\)

c, pthh: Zn+2HCl--->ZnCl2+H2 (2)

Theo pt(2): nHCl= 2.nH2= 2.0,3= 0,6 mol

=> mHCl= 0,6.36,5= 21,9 (g)

=> mddHCl= \(\dfrac{21,9.100\%}{10\%}=219\left(g\right)\)

nH2SO4=49/98=0,5(mol)

nAl=5,4/27=0,2(mol)

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

3________2

0,5_____0,2

Ta có: 0,5/3>0,2/2

=>H2SO4 dư

nH2SO4 dư=0,5-0,3=0,2(mol)

Theo pt: nH2=3/2nAl=3/2.0,2=0,3(mol)

=>VH2=0,3.22,4=6,72(l)

nZn=19,5/65=0,3(mol)

H2SO4+Zn--->ZnSO4+H2

1______1

0,2____0,3

Ta có: 0,1/1<0,3/1

=>Zn dư

Theo pt: nH2=nH2SO4=0,2(mol)

=>VH2=0,2.22,4=4,48(l)

=>\(\Sigma VH2=\)4,48+6,72=11,2(l)

Fe + 3/2Cl2 -to-> FeCl3

Fe + 2HCl --> FeCl2 + H2

3Fe + 2O2 -to-> Fe3O4

Fe + H2SO4 --> FeSO4 + H2

2Fe + 6H2SO4(đ) -to-> Fe2(SO4)3 + 3SO2 + 6H2O

Al + 3/2Cl2 -to-> AlCl3

2Al + 6HCl --> 2AlCl3 + 3H2

2Al + 3O2 -to-> 2Al2O3

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

2Al + 6H2SO4 (đ) -to-> Al2(SO4)3 + 3SO2 + 6H2O

Zn + Cl2 -to-> ZnCl2

Zn + 2HCl --> ZnCl2 + H2

Zn + 1/2O2 -to-> ZnO

Zn + H2SO4 --> ZnSO4 + H2

Zn + 2H2SO4(đ) -to-> ZnSO4 + SO2 + 2H2O

PTHH: 2Fe +3 Cl₂ \(\underrightarrow{t^o}\) 2FeCl₃
Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂\(\uparrow\)
Fe + H₂SO_{4 loãng} \(\rightarrow\) FeSO₄ + H₂\(\uparrow\)
2Fe + 6H₂SO_{4 đặc} \(\underrightarrow{t^o}\) Fe₂(SO₄)₃ + 6H₂O + 3SO₂\(\uparrow\)
2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂\(\uparrow\)
4Al + 3O₂ \(\underrightarrow{t^o}\) 2Al₂O₃
2Al + 6H₂SO_{4 đặc} \(\underrightarrow{t^o}\) Al₂(SO₄)₃ + 6H₂O + 3SO₂\(\uparrow\)
2Al + 3H₂SO_{4 loãng} \(\rightarrow\) Al₂(SO₄)₃ + 3H₂\(\uparrow\)
Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂\(\uparrow\)
Zn + H₂SO_{4 loãng} \(\rightarrow\) ZnSO₄ + H₂\(\uparrow\)
Zn + 2H₂SO_{4 đặc} \(\rightarrow\) ZnSO₄ + 2H₂O + SO₂\(\uparrow\)