\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2\left(tổng\right)}=n_{Fe}+n_{Zn}=0,2+0,2=0,4\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right);n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ m_{FeCl_2}=127.0,2=25,4\left(g\right)\)

\(n_{SO_2}=\dfrac{13,644}{22,4}=0,61\left(mol\right)\)

Đặt n _Fe = x (mol) =>\(m_{Fe}=56x\)

Vì m _Fe = m_Mg => \(n_{Mg}=\dfrac{56x}{24}=\dfrac{7}{3}x\)

n_Al = y(mol)

=> 56x + 56x + 27y = 16,14 (1)

\(Fe\rightarrow Fe^{3+}+3e\)                                       \(S^{+6}+2e\rightarrow S^{+4}\)

\(Mg\rightarrow Mg^{2+}+2e\)

\(Al\rightarrow Al^{3+}+3e\)

Bảo toàn e : 3x + \(\dfrac{7}{3}.2x\) + 3y = 0,61.2 (2)

Từ (1), (2) => x=0,12 ; y=0,1

=> m_Fe =m_Mg=0,12.56 = 6,72(g)

m _Al = 0,1.27=2,7(g)

\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\\n_{Fe}=c\left(mol\right)\end{matrix}\right.\)⇒ 24a + 27b + 56c = 26,05(1)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al +6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\)

\(Mg + Cl_2 \xrightarrow{t^o} MgCl_2\\ 2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ n_{Cl_2} = a + 1,5b + 1,5c = \dfrac{17,36}{22,4} = 0,775(3)\)

Từ (1)(2)(3) suy ra:  a = 0,325 ; b = -0,05 ; c = 0,35

→ Sai đề.

Chọn đáp án C

Gọi $n_{Fe} = a ; n_{Mg} = b; n_{Al} = c$

Ta có : 

$24b = 56a(1)$
$56a + 24b + 27c = 16,14(2)$

$n_{SO_2} = 0,61(mol)$
Bảo toàn electron : $3n_{Fe} + 2n_{Mg} + 3n_{Al} = 2n_{SO_2}$
$\Rightarrow 3a + 2b + 3c = 0,61.2(3)$

Từ (1)(2)(3) suy ra a = 0,12 ; b = 0,28 ; c = 0,1

$m_{Fe} = m_{Mg} = 0,12.56 = 6,72(gam)$

$m_{Al} = 0,1.27 = 2,7(gam)$

Đáp án : A

n_H2 = 0,25 mol; n_SO2 = 0,3 mol

2H⁺   + 2e → H₂                                  S⁺⁶ + 2e   → S⁺⁴

           0,5 <-- 0,25                                      0,6 <-- 0,3

n_Fe = 0,6 – 0,5 = 0,1 mol

=> m_Fe = 5,6g

Đáp án A

a) Gọi số mol Zn, Fe là a, b (mol)

=> 65a + 56b = 8,56 (1)

\(n_{H_2}=\dfrac{3,136}{22,4}=0,14\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           a--->2a-------->a----->a

            Fe + 2HCl --> FeCl₂ + H₂

           b----->2b------->b------>b

=> a + b = 0,14 (2)

(1)(2) => a = 0,08; b = 0,06 

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,08.65}{8,56}.100\%=60,748\%\\\%m_{Fe}=\dfrac{0,06.56}{8,56}.100\%=39,252\%\end{matrix}\right.\)

b) 

n_KOH = 0,2.0,1 = 0,02 (mol)

PTHH: KOH + HCl --> KCl + H₂O

           0,02-->0,02

=> n_HCl = 0,02 + 2a + 2b = 0,3 (mol)

=> \(C_{M\left(HCl\right)}=xM=\dfrac{0,3}{0,15}=2M\)

c) m = 0,08.136 + 0,06.127 = 18,5(g)

=> KL M là Kali (M=39, n=1)

Đáp án D