a)

Gọi $n_{Fe} = a(mol) ; n_{Al} =b (mol) \Rightarrow 56a + 27b = 11(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : $n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,1 ; b = 0,2

$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,9\%$
$\%m_{Al} = 100\% - 50,9\% = 49,1\%$

b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,8}{0,4} = 2M$

c)

$C_{M_{FeCl_2}} = \dfrac{0,1}{0,4} = 0,25M$
$C_{M_{AlCl_3}} =\dfrac{0,2}{0,4} = 0,5M$

\(a,n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ b,n_{H_2}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ c,n_{FeCl_2}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%\approx 22,93\%\)

Chọn B

Khi cho X tác dụng với HNO₃ thu được dung dịch T gồm Fe(NO₃)₂ (x); Fe(NO₃)₃ (y); Cu(NO₃)₂ (0,09).

Vậy C% Fe(NO₃)₃ = 5,67%

Đáp án B

a) 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

\(n_{CuSO_4}=0,1.1=0,1\left(mol\right)\)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: Fe + CuSO₄ --> FeSO₄ + Cu

Xét tỉ lệ \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) => CuSO₄ hết, Fe dư

PTHH: Fe + CuSO₄ --> FeSO₄ + Cu

_____0,1<---0,1---------->0,1

Fe + 2HCl --> FeCl₂ + H₂

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

=> V_H2 = 0,05.22,4 = 1,12(l)

b) \(C_{M\left(FeSO_4\right)}=\dfrac{0,1}{0,1}=1M\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH :

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

0,25      0,5                   0,25

\(a,m_{Fe}=0,25.56=14\left(g\right)\)

\(b,C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

______0,2_____0,4_____0,2 (mol)

a, \(m_{CuCl_2}=0,2.135=27\left(g\right)\)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{14,6}{300}.100\%\approx4,867\%\)

c, Ta có: m _{dd sau pư} = 16 + 300 = 316 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{27}{316}.100\%\approx8,54\%\)