a)

$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2o$

b)

Theo PTHH : $n_{Mg} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$m_{Mg} = 0,2.24 = 4,8(gam)$

$m_{MgO} = m_{hh} - m_{Mg} = 12,8 - 4,8 = 8(gam)$

c)

$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$
$n_{HCl} = 2n_{Mg} + 2n_{MgO} = 0,8(mol)$

$m_{dd\ HCl} = \dfrac{0,8.36,5}{14,6\%} = 200(gam)$

\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1          2            1           1

       0,2       0,4         0,2         0,2

      \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)

          1            6               2              3

         0,2         1,2            0,4

\(n_{Fe}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(m_{Fe2O3}=27,2-11,2=16\left(g\right)\)

⁰/₀Fe = \(\dfrac{11,2.100}{27,2}=41,18\)⁰/₀

⁰/₀Fe₂O₃ = \(\dfrac{16.100}{27,2}=58,82\)⁰/₀

b) Có : \(m_{Fe2O3}=16\left(g\right)\)

 \(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,4+1,2=1,6\left(mol\right)\)

\(V_{HCl}=\dfrac{1,6}{2}=0,8\left(l\right)\)

c) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

   \(n_{FeCl3}=\dfrac{1,2.2}{6}=0,4\left(mol\right)\)

  \(C_{M_{FeCl2}}=\dfrac{0,2}{0,8}=0,25\left(M\right)\)

  \(C_{M_{FeCl3}}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)

 Chúc bạn học tốt

\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)

\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}0,15(mol)\\ a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{Fe}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,15.56}{14,8}.100\%=56,76\%\\ \Rightarrow \%_{Cu}=100\%-56,76\%=43,24\%\\ c,n_{H_2SO_4}=0,15(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,15.98}{20\%}=73,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{73,5}{1,4}=52,5(l)\)

a) \(m_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

PTHH: MgCO₃ + H₂SO₄ --> MgSO₄ + CO₂ + H₂O

_____0,02<------0,02<-----------------0,02

=> m_MgCO3 = 0,02.84 = 1,68 (g)

=> \(\left\{{}\begin{matrix}\%MgCO_3=\dfrac{1,68}{10,3}.100\%=16,31\%\\\%MgCl_2=\dfrac{10,3-1,68}{10,3}.100\%=83,69\%\end{matrix}\right.\)

b) m_H2SO4 = 0,02.98 = 1,96 (g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{1,96}{100}.100\%=1,96\%\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,2     0,2               0,2       0,2

\(a,\%m_{Fe}=\dfrac{0,2.56}{20}.100\%=56\%\)

\(\%m_{Cu}=100\%-56\%=44\%\)

\(b,C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,1}=2\left(M\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

  \(1\)          \(1\)                             \(1\)

  \(0,2\)        \(0,2\)                      \(0,2\)

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(m_{Fe}=n.M=0,2.56=11,2\left(g\right)\)

\(^0/_0Fe=\dfrac{11,2}{20}.100^0/_0=56^0/_0\)

\(^0/_0Cu=100^0/_0-56^0/_0=44^0/_0\)

\(C_{M_{H_2SO_4}}=\dfrac{n}{V_{dd}}=\dfrac{0,2}{0,1}=2M\)

Ta có nH2 = 3,36/22,4 = 0,15 mol
        Fe +2 HCl -> FeCl2 + H2
       0,15.    0,3                   <-. 0,15.  ( Mol)
=> mFe = 0,15 × 56 = 8,4g
   => %Fe = 8,4/15×100% = 56% 
      => %Cu = 100% - 56% = 44%

=>VHCl =1\0,3=10\3 l

PTHH :  2Fe    +     6HCl   -->   2FeCl₃  +  3H₂   (1)

n_H2 =  \(\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

Từ (1) ->  n_Fe = \(\dfrac{2}{3}n_{H_2}=0.1\left(mol\right)\)

-> m_Fe = n.M = 0,1  . 56 = 5.6 (g) => %m_Fe = \(\dfrac{5.6}{15}x100\%\approx37.3\%\)

-> %m_Cu =  100% - 37.3% = 62.7 % 

a) Đặt: nZn=x(mol); nFe= y(mol) (x,y: nguyên, dương)

Zn + H2SO4 -> ZnSO4 + H2

x_______x_______x________x

Fe + H2SO4 -> FeSO4 + H2

y____y_________y___y(mol)

b) m(rắn)=mCu=3(g)

=> m(Zn, Fe)= 21,6 - 3= 18,6(g)

Ta có hpt:

\(\left\{{}\begin{matrix}65x+56y=18,6\\22,4x+22,4y=6,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

=> Zn= 65.0,2=13(g)

=>%mZn= (13/21,6).100=60,185%

%mCu=(3/21,6).100=13,889%

=>%mFe=25,926%

c) nH2SO4=x+y=0,3(mol) =>mH2SO4=29,4(g)

=> mddH2SO4= (29,4.100)/25=117,6(g)