\(n_{H_2}=\dfrac{2,36}{22,4}=\dfrac{59}{560}\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ n_{Al}=\dfrac{2}{3}.\dfrac{59}{560}=\dfrac{59}{840}\left(mol\right)\\ \Rightarrow\%m_{Al}=\dfrac{\dfrac{59}{840}.27}{50}.100\approx3,793\%\\ \Rightarrow\%m_{Al_2O_3}\approx96,207\%\)

a) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:       x                                  1,5x

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:       y                                 y

Ta có: \(\left\{{}\begin{matrix}27x+65y=24,9\\1,5x+y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}27x+65.\left(0,6-1,5x\right)=24,9\\y=0,6-1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

b, \(m_{Al}=0,2.27=5,4\left(g\right);m_{Zn}=24,9-5,4=19,5\left(g\right)\)

c) \(\%m_{Al}=\dfrac{5,4.100\%}{24,9}=21,69\%;\%m_{Zn}=100\%-21,69\%=78,31\%\)

d) 

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,2     0,6          0,2       0,3

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:      0,3     0,6        0,3       0,3

\(m_{ddHCl}=\dfrac{\left(0,6+0,6\right).36,5.100}{14}=312,857\left(g\right)\)

e) m_{dd sau pứ} = 24,9 + 312,857 - (0,3+0,3).2 = 336,557 (g)

\(C\%_{ddAlCl_3}=\dfrac{0,2.133,5.100\%}{336,557}=7,93\%\)

\(C\%_{ddZnCl_2}=\dfrac{0,3.136.100\%}{336,557}=12,12\%\)

a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)

              1           2              1            1

              a           2a

             \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

                   1            6              2              3

                   b           6b

b) Gọi a là số mol của CuO

           b là số mol của Fe₂O₃

\(m_{CuO}+m_{Fe2O3}=25\left(g\right)\)

⇒ \(n_{CuO}.M_{CuO}+n_{Fe2O3}.M_{Fe2O3}=25g\)

 ⇒ 80a + 160b = 25g (1) 

Ta có : 200ml = 0,2l

\(n_{HCl}=3,5.0,2=0,7\left(mol\right)\)

⇒ 2a + 6b = 0,7(2)

Từ (1),(2) ta có hệ phương trình :

      80a + 160b = 25g

       2a + 6b = 0,7

    ⇒ \(\left\{{}\begin{matrix}a=0,2375\\b=0,0375\end{matrix}\right.\)

\(m_{CuO}=0,2375.80=19\left(g\right)\)

\(m_{Fe2O3}=0,0375.160=6\left(g\right)\)

 Chúc bạn học tốt

\(n_{H_2}=\dfrac{0.56}{22.4}=0.025\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{2}{3}\cdot0.025=\dfrac{1}{60}\left(mol\right)\)

\(m_{Al}=\dfrac{1}{60}\cdot27=0.45\left(g\right)\)

\(m_{Cu}=25-0.45=24.55\left(g\right)\)

\(\%Cu=\dfrac{24.55}{25}\cdot100\%=98.2\%\)

\(\%Al=100-98.2=1.8\%\)

\(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

\(2Al+6H_2SO_{4\left(đ\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

\(n_{Cu}=\dfrac{24.55}{64}=\dfrac{491}{1280}\left(mol\right)\)

\(V_{SO_2}=\left(\dfrac{1}{60}\cdot\dfrac{3}{2}+\dfrac{491}{1280}\right)\cdot22.4=9.1525\left(l\right)\)

a)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
b)

$n_{H_2} = \dfrac{0,56}{22,4} = 0,225(mol)$

Theo PTHH : $n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{60}(mol)$
$m_{Al} = \dfrac{1}{60}.27 = 0,45(gam)$
$m_{Cu} = 25 - 0,45 = 24,55(gam)$

c)

$\%m_{Al} = \dfrac{0,45}{25}.100\% = 1,8\%$
$\%m_{Cu} = 100\% -1,8\% = 98,2\%$

d)

$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O$

Theo PTHH : 

$n_{SO_2} = n_{Cu} + \dfrac{3}{2}n_{Al} = \dfrac{24,55}{64} + \dfrac{1}{60}.\dfrac{3}{2} = 0,41(mol)$
$V_{SO_2} = 0,41.22,4 = 9,184(lít)$

2Al+6HCl->2AlCl3+ 3H2

n(H2)=0,025mol

n(al)= 1/60 mol

m(Al)= 0,45g

m(Cu)= 25-0,45=24,55g

%m(Cu)=98,2%

%m(Al)=1,8%

Cu+2H2SO4(đ)-> CuSO4+SO2+H2O

2Al +6H2SO4(đ)-> Al2(SO4)3 + 3SO2+6H2O

n(so2)=n(Cu)+1,5n(al)= 24,55/64 +1,5*1/60 =523/1280 mol

V(so2)=9,1525lit

\(n_{H_2}=\dfrac{2,464}{22,4}=0,11mol\)

\(\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\Rightarrow Muối\left\{{}\begin{matrix}Al_2\left(SO_4\right)_3\\FeSO_4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}BTe:3x+2y=2n_{H_2}=0,22\\\dfrac{x}{2}\cdot342+y\cdot152=14,44\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04mol\\y=0,05mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,04\cdot27=1,08g\\m_{Fe}=0,05\cdot56=2,8g\end{matrix}\right.\)

\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\downarrow\)

0,02                                                   0,06

\(FeSO_4+BaCl_2\rightarrow BaSO_4\downarrow+FeCl_2\)

0,05                          0,05

\(\Rightarrow\Sigma n_{\downarrow}=0,06+0,05=0,11\Rightarrow m_{BaSO_4}=x=25,63g\)

Fe+H2SO4->FeSO4+H2

0,15-----------------------0,15 mol

n H2=0,15 mol

=>m Fe=0,15.56=8,4g

=>m Cu=11,8-8,4=3,4g

n Al = 5,4 / 27 =0,2 mol

PTHH:

2 Al + 3 H2SO4 = Al2(SO4)3 + 3 H2

Theo pthh:

n H2 = 3/2 nAl = 3/2 * 0,2 = 0,3 mol.

V H2 = 0,3 * 22,4 = 6,72 lít.

=>m muối = 0,1.342=34,2g

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Gộp cả phần a và b

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2}=0,3mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)

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