a. B gồm AlCl₃ 

\(b.n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,4      1,2             0,4           0,6

\(\%m_{Al}=\dfrac{0,4.27}{23,6}\cdot100\%=45,76\%\\ \%m_{Cu}=100\%-45,76=54,24\%\\ c.m_B=m_{AlCl_3}=1,2.133,5=106,2g\)

a)

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$n_{H_2} = \dfrac{13,44}{22,4} = 0,6(mol)$

Theo PTHH : 

$n_{Al} = \dfrac{2}{3}n_{H_2} = 0,4(mol)$

$m_{Al} = 0,4.27 = 10,8(gam)$
$m_{Cu} = 20,4 - 10,8 = 9,6(gam)$

b) $n_{H_2SO_4} = n_{H_2} = 0,6(mol)$
$m = \dfrac{0,6.98}{14,7\%} = 400(gam)$

Bài 1:

\(n_{H_2SO_4}=\dfrac{200.14,7\%}{98}=0,3\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=342.0,1=34,2\left(g\right)\)

Bài 2:

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{Mg}=0,15.24=3,6\left(g\right)\\ \%m_{Mg}=\dfrac{3,6}{10}.100=36\%\\ \%m_{Cu}=100\%-36\%=64\%\)

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

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

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

=> m_Cu = 10-0,1.56 = 4,4(g)

=> \(\%Cu=\dfrac{4,4}{10}.100\%=44\%\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ m_{Fe}=0,1.56=5,6\left(g\right)\\ \%m_{Fe}=\dfrac{5,6}{8}.100=70\%\\ \Rightarrow\%m_{Cu}=100\%-70\%=30\%\)

\(n_k=0,6mol\)

\(d_k=d_{H_2}=26\Rightarrow d_k=52\)

Áp dụng sơ đồ đường chéo tính được là: \(n_{SO_2}=0,2mol;n_{NO_2}=0,4mol\)

Bảo toàn e: \(5^{+6}+2e\rightarrow S^{+4}\)

mol: \(0,4\rightarrow0,2\)

\(N^{+5}+1e\rightarrow N^{+4}\)

mol: \(0,4\rightarrow0,4\)

\(m_{\text{muối}}=m_{kl}+m_{SO_4^{2^-}}+m_{NO_3^-}=55,8g\)

Ta có : C1=2C2

=> Gọi n_H2SO4 =x 

=> n _HCl = 2x

Bảo toàn nguyên tố H :\(n_{HCl}.1+n_{H_2SO_4}.2=n_{H_2}.2\)

\(\Rightarrow2a+2a=\dfrac{13,44}{22,4}=0,6.2\)

=>a = 0,3(mol)

=> CM_HCl = \(\dfrac{0,6}{0,3}=2M\); CM_H2SO4 = \(\dfrac{0,3}{0,3}=1M\)

Dung dịch B gồm : Mg ²⁺ , Al³⁺ , Cl^- , SO4 ^2-

\(n_{Cl^-}=n_{HCl}=0,6\left(mol\right);n_{SO_4^{2-}}=n_{H_2SO_4}=0,3\left(mol\right)\)

Bảo toàn điện tích cho dung dịch B:

\(n_{Mg}.2+n_{Al}.3=0,6+0,3.2\) (1)

Theo đề bài : \(24.n_{Mg}+27.n_{Al}=12,6\) (2)

Từ (1), (2)=> \(\left\{{}\begin{matrix}n_{Mg}=0,3\\n_{Al}=0,2\end{matrix}\right.\)

=> \(\%m_{Mg}=\dfrac{0,3.24}{12,6}.100=57,14\%\)

=> % m _Al = 100 -57.14 = 42,86%