\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1         2              1          1

       0,1     0,2            0,1       0,1

a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

100ml = 0,1l

\(C_{M_{ddHCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)

c) \(n_{FeCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(C_{M_{FeCl2}}=\dfrac{0,1}{2}=0,05\left(M\right)\)

 Chúc bạn học tốt

Mình xin lỗi bạn nhé , bạn sửa lại giúp mình chỗ : 

\(C_{M_{FeCl2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{13}{65}=0,2mol\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\\ c)n_{HCl}=0,2.2=0,4mol\\ C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ d)m_{ZnCl_2}=0,2.136=27,2g\)

a)

$n_{Al} = \dfrac{8,1}{27} = 0,3(mol)$

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

b) $n_{HCl} = 3n_{Al} = 0,9(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,9.36,5}{3,65\%} = 900(gam)$

c)

$m_{dd\ sau\ pư}= 8,1 + 900 - 0,45.2 = 907,2(gam)$

$n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,15(mol)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{907,2}.100\% = 5,65\%$

a)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$

$MgO + 2HCl \to MgCl_2 + H_2O$

Theo PTHH : $n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)$
$\%m_{Al} = \dfrac{0,2.27}{18,8}.100\% = 28,7\%$

$\%m_{MgO} = 100\% - 28,7\% =71,3\%$

b) $n_{MgO} = 0,335(mol)$

Theo PTHH : $n_{HCl} = 2n_{H_2} + 2n_{MgO} =1,27(mol)$

$V_{dd\ HCl} = \dfrac{1,27}{1,6} = 0,79375(lít)$

c) 

$H_2 + O_{oxit} \to H_2O$

$\Rightarrow n_{O(oxit)} = n_{H_2} = 0,3(mol)$
$\Rightarrow n_{Fe} = \dfrac{17,4 - 0,3.16}{56} = 0,225(mol)$

Ta có : 

$n_{Fe} : n_O = 0,225 : 0,3 = 3 : 4$

Vậy oxit là $Fe_3O_4$

1)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

____0,1----->0,15

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)

2)

\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

_______0,2------------------------------>0,2

=> V_CO2 = 0,2.22,4 = 4,48(l)

3)

\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)

PTHH: 2A + Cl₂ --t^o--> 2ACl

____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)

=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)

4)

n_HCl = 0,2.3 = 0,6(mol)

PTHH: M + 2HCl --> MCl₂ + H₂

____0,3<-----0,6

=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)

nFe=0,1 mol

Fe              +2HCl=>FeCl2+H2

0,1 mol=>0,2 mol          =>0,1 mol

VH2=0,1.22,4=2,24 lít

nHCl=0,2 mol=>mHCl=0,2.36,5=7,3g

=>C% dd HCl=7,3/200.100%=3,65%

a ,\(Zn+2HCl=>ZnCl_2+H_2\) (1)

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

theo (1) \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\) 

=> \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)  

c, Theo (1) \(n_{HCl}=2n_{Zn}=0,2\left(mol\right)\)  

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

nồng độ % dung dịch axit đã dùng là

\(\frac{7,3}{200}.100\%=36,5\%\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

Câu c và câu d không liên quan tới dữ liệu đề bài cho !

c, d vẫn tính đc mà

a) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

           0,05---->0,1----->0,05--->0,05

=> \(C\%\left(HCl\right)=\dfrac{0,1.36,5}{150}.100\%=2,433\%\)

b) \(C\%\left(CuCl_2\right)=\dfrac{0,05.135}{4+150}.100\%=4,383\%\)