a, \(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

b, \(n_{CH_3COOH}=0,1.2=0,2\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, Theo PT: \(n_{Zn}=\dfrac{1}{2}n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,5}.100\%\approx68,42\%\\\%m_{Cu}\approx31,58\%\end{matrix}\right.\)

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

PTHH:

`2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2`

`Fe + H_2SO_4 -> FeSO_4 + H_`

b) `n_{H_2} = (0,56)/(22,4) = 0,025 (mol)`

Theo PT: `n_{H_2} = n_{Fe} + 3/2 n_{Al}`

`=> b + 1,5a = 0,025`

Giải hpt \(\left\{{}\begin{matrix}27a+56b=0,83\\1,5a+b=0,025\end{matrix}\right.\Leftrightarrow a=b=0,01\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)

a) PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

_____0,02<---0,03<---------------------0,03

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,16}.100\%=25\%\\\%Cu=100\%-25\%=75\%\end{matrix}\right.\)

c) m_H2SO4 = 0,03.98 = 2,94 (g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{2,94}{200}.100\%=1,47\%\)

\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 16,8 - 6,4 = 10,4(1)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)\)

Từ (1)(2) suy ra: a = 0,1 ; b = 0,2

Vậy :

\(\%m_{Fe} = \dfrac{0,1.56}{16,8}.100\% = 33,33\%\\ \%m_{Mg} = \dfrac{0,2.24}{16,8}.100\% = 28,57\%\\ \%m_{Cu} = 100\% - 33,33\% - 28,57\% = 38,1\%\)

Bài 1:

a+b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{Fe}\)

\(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\) \(\Rightarrow m_{Cu}=6,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{17,6}\cdot100\%\approx63,64\%\\\%m_{Cu}=36,36\%\end{matrix}\right.\)

c) Ta có: \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

Bảo toàn nguyên tố: \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)=n_{CuSO_4}\)

\(\Rightarrow m_{muối}=0,1\cdot400+0,1\cdot160=56\left(g\right)\)

Bài 2:

Quy đổi hh gồm Fe (a mol) và O (b mol)

\(\Rightarrow56a+16b=27,6\)  (1)

Ta có: \(n_{SO_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

Bảo toàn electron: \(3n_{Fe}=2n_O+2n_{SO_2}\) \(\Rightarrow3a-2b=0,45\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,39\\b=0,36\end{matrix}\right.\)

Bảo toàn nguyên tố: \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,195\left(mol\right)\) \(\Rightarrow m_{Fe_2\left(SO_4\right)_3}=0,195\cdot400=78\left(g\right)\)

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

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:       x                                                     1,5x

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:      y                                                 y

Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)

b) 

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:      0,1      0,15                  0,05                            

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:     0,1       0,1                 0,1

\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)

m_{dd sau pứ} = 5,1+250-0,15.2 = 254,8(g)

\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)

\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)

\(Đặt:\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Cu}=y\left(mol\right)\end{matrix}\right.\\ Fe\rightarrow Fe^{3+}+3e\\ Cu\rightarrow Cu^{2+}+2e\\ 4H^++NO_3^-+3e\rightarrow NO+2H_2O\\ Bảotoàne:3x+2y=0,4.3\\ Tacó:56x+64y=30,4\\ \Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=36,84\%\\\%m_{Cu}=63,16\%\end{matrix}\right.\\ n_{HNO_3}=4n_{NO}=0,4.4=1,6\left(mol\right)\\ \Rightarrow V_{HNO_3}=\dfrac{1,6}{1}=1,6\left(l\right)\\ m_{muối}=m_{Fe\left(NO_3\right)_3}+m_{Cu\left(NO_3\right)_2}=0,2.242+0,3.188=104,8\left(g\right)\)

a,\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:      x                                 x    

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:      y                                 y

Ta có: \(\left\{{}\begin{matrix}65x+56y=30,7\\x+y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Zn}=\dfrac{0,3.65.100\%}{30,7}=63,52\%;\%m_{Fe}=100\%-63,52\%=36,48\%\)

b,

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,3     0,6                             

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,2     0,4

n_HCl = 0,6+0,4 = 1 (mol)

\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ a)ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,15    0,15            0,15          0,15

\(b)m_{Fe}=0,15.56=8,4g\\ m_{ZnO}=16,5-8,4=8,1g\\ c)n_{ZnO}=\dfrac{8,1}{81}=0,1mol\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

0,1          0,1               0,1            0,1

\(V_{ddH_2SO_4}=\dfrac{0,15+0,1}{2}=0,125M\\ d)Fe+CuSO_4\rightarrow FeSO_4+Cu\)

0,15        0,15            0,15          0,15

\(m_{rắn}=m_{ZnO}+m_{Cu}=8,1+0,15.64=17,7g\)