Trong \(20,4g\) hỗn hợp có: \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow65a+56b+27c=20,4\left(1\right)\)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45mol\)

\(BTe:2n_{Zn}+2n_{Fe}+3n_{Al}=2n_{H_2}\)

\(\Rightarrow2a+2b+3c=2\cdot0,45\left(2\right)\)

Trong \(0,2mol\) hhX có \(\left\{{}\begin{matrix}Zn:ka\left(mol\right)\\Fe:kb\left(mol\right)\\Al:kc\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow ka+kb+kc=0,2\)

\(n_{Cl_2}=\dfrac{6,16}{22,4}=0,275mol\)

\(BTe:2n_{Zn}+3n_{Fe}+3n_{Al}=2n_{Cl_2}\)

\(\Rightarrow2ka+3kb+3kc=2\cdot0,275\)

Xét thương:

 \(\dfrac{ka+kb+kc}{2ka+3kb+3kc}=\dfrac{0,2}{2\cdot0,275}\Rightarrow\dfrac{a+b+c}{2a+3b+3c}=\dfrac{4}{11}\)

\(\Rightarrow3a-b-c=0\left(3\right)\)

Từ (1), (2), (3)\(\Rightarrow\left\{{}\begin{matrix}a=0,1mol\\b=0,2mol\\c=0,1mol\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Zn}=6,5g\\m_{Fe}=11,2g\\m_{Al}=2,7g\end{matrix}\right.\)

a)

\(n_{H_2\left(1\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂ (1)

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

=> m_Na = 0,6.23 = 13,8 (g)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,1<-0,2

=> m_Fe = 0,1.56 = 5,6 (g)

m_Cu = 10 (g)

\(\left\{{}\begin{matrix}\%Na=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%Fe=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%Cu=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)

PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

             \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=\dfrac{17,4}{\dfrac{0,3}{y}}=58y\left(g/mol\right)\)

=> 56x = 42y

=> \(\dfrac{x}{y}=\dfrac{3}{4}\) => CTHH: Fe₃O₄

a)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

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

             Fe + 2HCl --> FeCl₂ + H₂

            0,1<--0,2

=> \(\left\{{}\begin{matrix}m_{Na}=0,6.23=13,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Cu}=10\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%m_{Fe}=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%m_{Cu}=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)
PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

            \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=56x+16y=\dfrac{17,4}{\dfrac{0,3}{y}}\left(g/mol\right)\)

=> \(\dfrac{x}{y}=\dfrac{3}{4}\)

=> CTHH: Fe₃O₄

a, \(n_{H_2}=0,09\left(mol\right)\)

BT e, có: 2n_X = 2n_H2 ⇒ n_X = 0,09 (mol)

\(\Rightarrow\overline{M}_X=\dfrac{2,64}{0,09}=29,33\left(g/mol\right)\)

Mà: A, B thuộc 2 chu kì liên tiếp.

→ Mg và Ca.

Ta có: \(\left\{{}\begin{matrix}24n_{Mg}+40n_{Ca}=2,64\\2n_{Mg}+2n_{Ca}=0,09.2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,06\left(mol\right)\\n_{Ca}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,06.24}{2,64}.100\%\approx54,55\%\\\%m_{Ca}\approx45,45\%\end{matrix}\right.\)

b, BTNT H, có: \(n_{H_2SO_4}=n_{H_2}=0,09\left(mol\right)\Rightarrow V_{ddH_2SO_4}=\dfrac{0,09}{2}=0,045\left(l\right)\)

BTNT Mg: n_MgSO4 = n_Mg = 0,06 (mol)

\(\Rightarrow C_{M_{MgSO_4}}=\dfrac{0,06}{0,045}=\dfrac{4}{3}\left(M\right)\)

a) Đặt  số mol của MO, M(OH)₂, MCO₃ tương ứng là x, y, z.

Nếu tạo muối trung hòa ta có các phản ứng:

MO  +  H₂SO₄   →MSO₄    +   H₂O                                   (1)

M(OH)₂  +  H₂SO₄    →MSO₄    +  2H₂O                          (2)

MCO₃   +  H₂SO₄    →MSO₄    +   H₂O + CO₂              (3)

Nếu tạo muối axít ta có các phản ứng:

MO  +  2H₂SO₄    →M(HSO₄)₂   +   H₂O                         (4)

M(OH)₂  +  2H₂SO₄    →M(HSO₄)₂      +  2H₂O                (5)

MCO₃   +  2H₂SO₄   →M(HSO₄)₂  +   H₂O + CO₂                             (6)

Ta có : 

– TH₁: Nếu muối là MSO₄   M + 96 = 218   M = 122 (loại)

– TH₂: Nếu là muối M(HSO₄)₂   M + 97.2 = 218  M = 24 (Mg)

Vậy xảy ra phản ứng (4, 5, 6) tạo muối Mg(HSO₄)₂                                            

b) Theo (4, 5, 6)    Số mol CO₂ = 0,448/22,4 = 0,02 molz = 0,02  (I)

2x + 2y + 2z = 0,12             (II)

Đề bài:       40x + 58y + 84z = 3,64 (III) 

Giải hệ (I, II, III): x = 0,02; y = 0,02; z = 0,02

%MgO = 40.0,02.100/3,64 = 21,98%

%Mg(OH)₂ = 58.0,02.100/3,64 = 31,87%   

%MgCO₃ = 84.0,02.100/3,64 = 46,15%

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

              a_______a_______a_____a    (mol)

            \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

                2b______3b__________b_____3b    (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27\cdot2b=11\\a+3b=0,2\cdot2=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{11}\cdot100\%\approx50,91\%\\\%m_{Al}=49,09\%\end{matrix}\right.\)

Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{FeSO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\) 

a) nH2SO4=0,4(mol)

Đặt: nFe=x(mol); nAl=y(mol) (x,y>0)

PTHH: Fe + H2SO4 -> FeSO4 + H2

x________x______x______x(mol)

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

y____1,5y_______0,5y_______1,5y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=> mFe=0,1.56=5,6(g)

=>%mFe=(5,6/11).100=50,909%

=>%mAl= 49,091%

b) V(H2,đktc)=0,4.22,4=8,96(l)

c) nAl2(SO4)3= 0,5y=0,5.0,2=0,1(mol)

nFeSO4=x=0,1(mol)

Vddsau=VddH2SO4=0,2(l)

=>CMddAl2(SO4)3= 0,1/0,2=0,5(M)

CMddFeSO4=0,1/0,2=0,5(M)