\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right),n_{Al}=c\left(mol\right),n_{Zn}=d\left(mol\right)\)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(\)\(BTe:\)

\(2a+2b+3c+2d=0.2\left(1\right)\)

\(n_{Cl_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(BTe:\)

\(3a+2b+3c+2c=0.15\cdot2=0.3\left(2\right)\)

\(\left(2\right)-\left(1\right):a=0.3-0.2=0.1\)

\(\%Fe=\dfrac{0.1\cdot56}{32}\cdot100\%=17.5\%\)

Gọi a, b, c, d là mol mỗi chất trong 32g X 

$n_{H_2}=0,1mol$

Bảo toàn e: $2a+2b+3c+2d=0,1.2=0,2$ (1) 

$n_{S{O}_2}=0,15mol$

Bảo toàn e: $3a+2b+3c+2d=0,15.2=0,3$  (2)

Lấy (2) trừ (1) => $a=0,1$

$\%Fe=\frac{0,\mathrm{1.56.100}}{32}=17,5\%$

Gọi a, b, c, d là mol mỗi chất trong 32g X 

$n_{H_2}=0,1mol$

Bảo toàn e: $2a+2b+3c+2d=0,1.2=0,2$ (1) 

$n_{S{O}_2}=0,15mol$

Bảo toàn e: $3a+2b+3c+2d=0,15.2=0,3$  (2)

Lấy (2) trừ (1) => $a=0,1$

$\%Fe=\frac{0,\mathrm{1.56.100}}{32}=17,5\%$

Gọi: \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\\n_{Zn}=d\left(mol\right)\end{matrix}\right.\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

BT e, có: 2n_Fe + 2n_Mg + 3n_Al + 2n_Zn = 2n_H2

⇒ 2a + 2b + 3c + 2d = 0,2 (1)

Ta có: \(n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

BT e, có: 3n_Fe + 2n_Mg + 3n_Al + 2n_Zn = 2n_SO2

⇒ 3a + 2b + 3c + 2d = 0,3 (2)

Trừ 2 vế của (2) và (1), được: a = 0,1 (mol)

\(\Rightarrow\%m_{Fe}=\dfrac{0,1.56}{32}.100\%=17,5\%\)

Chọn đáp án A

Gọi $n_{Fe} = a(mol), n_{Zn} = b(mol) , n_{Al} = c(mol) \Rightarrow 56a + 65b + 27c = 20,4(1)$

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$2Al +3 H_2SO_4 \to Al_2(SO_4)_3 +3 H_2$

Theo PTHH : $n_{H_2} = a + b + 1,5c = \dfrac{10,08}{22,4} = 0,45(mol)(2)$

Mặt khác : 

$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$Zn + Cl_2 \xrightarrow{t^o} ZnCl_2$
$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$

Theo PTHH : $n_{Cl_2} = 1,5n_{Fe} + n_{Zn} + 1,5n_{Al}$

Suy ra : \dfrac{1,5a + b + 1,5c}{a + b + c} = \dfrac{0,275}{0,2}(3)$

Từ (1)(2)(3) suy ra : a = 0,2 ; b = 0,1 ; c = 0,1

$\%m_{Fe} = \dfrac{0,2.56}{20,4}.100\% = 54,9\%$

$\%m_{Zn} = \dfrac{0,1.65}{20,4}.100\% = 31,9\%$
$\%m_{Al} = 100\% - 54,9\% - 31,9\% = 13,2\%$

a) 

TN1: Gọi (n_Zn; n_Fe; n_Cu) = (a; b; c)

=> 65a + 56b + 64c = 18,5 (1)

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            a---------------------->a

            Fe + 2HCl --> FeCl₂ + H₂

             b----------------------->b

=> a + b = 0,2 (2)

TN2: Gọi (n_Zn; n_Fe; n_Cu) = (ak; bk; ck)

=> ak + bk + ck = 0,15 (3)

PTHH: Zn + Cl₂ --t^o--> ZnCl₂

           ak-->ak

           2Fe + 3Cl₂ --t^o--> 2FeCl₃

            bk--->1,5bk

           Cu + Cl₂ --t^o--> CuCl₂

           ck-->ck

=> \(ak+1,5bk+ck=\dfrac{3,92}{22,4}=0,175\)(4)

(1)(2)(3)(4) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\\c=0,1\left(mol\right)\\k=0,5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{18,5}.100\%=35,135\%\\\%m_{Fe}=\dfrac{0,1.56}{18,5}.100\%=30,27\%\\\%m_{Cu}=\dfrac{0,1.64}{18,5}.100\%=34,595\%\end{matrix}\right.\)

b) n_O(oxit) = \(\dfrac{23,7-18,5}{16}=0,325\left(mol\right)\)

=> n_H2O = 0,325 (mol)

=> n_HCl = 0,65 (mol)

=> \(V=\dfrac{0,65}{1}=0,65\left(l\right)=650\left(ml\right)\)

Đáp án : A

X + Cl₂ :

Bảo toàn khối lượng : m_Cl2 = m_muối – m_KL = 31,95g

3n_Fe + 3n_Cr + 3n_Al = 2n_Cl2 = 0,9 mol

X + HCl : 2n_Fe + 2n_Cr + 3n_Al = 2n_H2 = 0,7 mol

=> n_Al  = 0,1 mol ; n_Fe + n_Cr  = 0,2 mol

=> %n_Al = 33,33%

Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Zn}=c\left(mol\right)\\n_{Al}=d\left(mol\right)\end{matrix}\right.\) \(\Rightarrow95a+127b+136c+133,5d=40,45\)  (1)

Sau p/ứ với Clo, ta được: \(95a+162,5b+136c+133,5d=44\)  (2)

Lấy PT (2) trừ PT (1) \(\Rightarrow35,5b=3,55\) \(\Rightarrow b=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{0,1\cdot56}{13,47}\cdot100\%\approx41,57\%\)