Chọn đáp án C

a) Y là Cu

$m_{Cu} = 8(gam)$

Gọi $n_{Al} = a(mol) ; n_{Fe} = b(mol)$

Ta có : $27a + 56b + 8 = 13,45(1)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2} = 1,5a + b = \dfrac{5,6}{22,4} = 0,25(2)$

Từ (1)(2) suy ra  a = 0,15 ; b = 0,025$

$\%m_{Cu} = \dfrac{8}{13,45}.100\% = 59,47\%$
$\%m_{Al} = \dfrac{0,15.27}{13,45}.100\% = 30,11\%$

$\%m_{Fe} = 10,42\%$

b)

$n_{H_2SO_4} = n_{H_2} = 0,25(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,25}{0,5} = 0,5(lít)$

Chọn A

Đáp án A

n H 2   =   0 , 2   ( m o l )

=> m_hh= m_Fe + m_Al

Bảo toàn electron:

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,25}.100\%\approx51,43\%\\\%m_{Al_2O_3}\approx48,57\%\end{matrix}\right.\)

b, \(n_{Al_2O_3}=\dfrac{5,25-0,1.27}{102}=0,025\left(mol\right)\)

Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=0,45\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,45.36,5}{29,2\%}=56,25\left(g\right)\)

c, \(n_{H_2SO_4}=\dfrac{1}{2}n_{HCl}=0,225\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,225.98}{19,6\%}=112,5\left(g\right)\)

\(n_{Al}=a\left(mol\right)\)

\(n_{Fe}=b\left(mol\right)\)

\(m=27a+56b=19.3\left(g\right)\left(1\right)\)

\(n_{H^+}=0.2\cdot2+0.2\cdot2.25\cdot2=1.3\left(mol\right)\)

\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)

\(Fe+2H^+\rightarrow Fe^{2+}+H_2\)

\(n_{H^+}=3a+2b=1.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.3,b=0.2\)

\(\%Al=\dfrac{0.3\cdot27}{19.3}\cdot100\%=41.96\%\)

\(\%Fe=58.04\%\)

\(b.\)

\(n_{H_2}=\dfrac{1}{2}n_{H^+}=0.65\left(mol\right)\)

Bảo toàn khối lượng : 

\(m_{Muối}=19.3+0.4\cdot36.5+0.45\cdot98-0.65\cdot2=76.7\left(g\right)\)

em cảm ơn ạ!