a) $Fe + 2HCl \to FeCl_2 + H_2$

 Theo PTHH : n H2 = n Fe = 8,4/56 = 0,15(mol)

V H2 = 0,15.22,4 = 3,36(lít)

b) n HCl = 2n Fe = 0,3(mol)

=> CM HCl = 0,3/0,2 = 1,5M

c) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 32/80 = 0,4 > n H2 = 0,15 mol nên CuO dư

Theo PTHH : n Cu = n H2 = 0,15 mol

=> m Cu = 0,15.64 = 9,6 gam

\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)

\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ n_{Al}=0,4(mol)\\ a/\\ n_{H_2}=\frac{3}{2}.0,4=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,6(mol)\\ n_{ddH_2SO_4}=\frac{0,6.98.100}{20}=294(g)\\ c/\\ n_{Al_2(SO_4)_3}=0,2(mol)\\ C\%_{Al_2(SO_4)_3}=\frac{0,2.342}{10,8+294-0,6.2}.100\%=22,52\%\)

`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`

 `0,1`      `0,2`          `0,1`       `0,1`         `(mol)`

`n_[Fe]=[5,6]/56=0,1(mol)`

`a)V_[H_2]=0,1.22,4=2,24(l)`

`b)C_[M_[HCl]]=[0,2]/[0,1]=2(M)`

Fe+H2SO4->FeSO4+H2

0,15---0,15-----0,15---0,15 mol

n Fe=8,4\56=0,15 mol

=>VH2=0,15.22,4=3,36l

=>m H2SO4=0,15.98=14,7g

=>C% H2SO4=14,7\245 .100=6%

=>m dd muối=8,4+245-0,15.2=253,1g

=>C% muối =0,15.152\253,1 .100=9%

Bài 1

Bài 5

Fe + 2CH₃COOH \(\rightarrow\) (CH₃COO)₂Fe + H₂(1)

nCH₃COOH = \(\dfrac{4,5}{60}=0,075mol\)

a) THeo pt: n(CH₃COO)₂Fe = \(\dfrac{1}{2}.nCH_3COOH=0,0375mol\)

=> m = 6,525g

c) Theo pt (1) nH₂ = 1/2nCH₃COOH = 0,0375 mol

2H₂ + O₂ \(\xrightarrow[]{t^o}\) 2H₂O

Theo pt: nO₂ = 0,5nH₂ = 0,01875mol

=> VO₂ = 0,42 lít

=> Vkk = 0,42.5 = 2,1 lít

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

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

           0,1--->0,2------->0,1----->0,1

V_H2 = 0,1.22,4 = 2,24 (l)

b, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)

c, \(C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\)

1)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,15->0,3--->0,15-->0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

c) m_{dd sau pư} = 8,4 + 250 - 0,15.2 = 258,1 (g)

=> \(C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%=7,38\%\)

2)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4---->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

m_ZnCl2 = 0,2.136 = 27,2 (g)

c) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,2}=2M\)

d) 

PTHH: A + 2HCl --> ACl₂ + H₂

           0,2<--0,4

=> \(M_A=\dfrac{4,8}{0,2}=24\left(g/mol\right)\)

=> A là Mg(Magie)