a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+65b=12,1\)  (1)

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

Bảo toàn electron: \(2n_{Fe}+2n_{Zn}=2n_{H_2}\) \(\Rightarrow2a+2b=0,4\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{12,1}\cdot100\%\approx46,28\%\\\%m_{Zn}=53,72\%\end{matrix}\right.\)

b)

Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=n_{Zn}=n_{ZnSO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(p.ứ\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow\Sigma n_{H_2SO_4}=0,2\cdot110\%=0,22\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1\cdot152=15,2\left(g\right)\\m_{ZnSO_4}=0,1\cdot161=16,1\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,22-0,2\right)\cdot98=1,96\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{KL}+m_{ddH_2SO_4}-m_{H_2}=211,7\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{15,2}{211,7}\cdot100\%\approx7,18\%\\C\%_{ZnSO_4}=\dfrac{16,1}{211,7}\cdot100\%\approx7,61\%\\C\%_{H_2SO_4}=\dfrac{1,96}{22,4}\cdot100\%\approx0,93\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)

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

0,1                           0,05           0,15

\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)

\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)

a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)

\(\%m_{Ag}=100\%-64,28\%=35,72\%\)

b)\(m_{muối}=0,05\cdot342=17,1g\)

a) Sửa đề: dd H₂SO₄ 9,8%

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)

\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)

b) Tương tự câu a

a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (1)

                \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)  (2)

b) Ta có: \(\Sigma n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi số mol của Mg là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=a\)

Gọi số mol của Zn là b \(\Rightarrow n_{H_2\left(2\right)}=b\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}a+b=0,4\\24a+65b=17,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2\cdot24}{17,8}\cdot100\%\approx26,97\%\\\%m_{Zn}\approx73,03\%\end{matrix}\right.\)

Chất không tan là Ag.

=> mAg= 6,25(g)

nH2=0,25(mol)

PTHH: Zn + H2SO4 -> ZnSO4 + H2

-> nZn=nH2= 0,25(mol)

=>mZn= 0,25 . 65=16,25(g)

=> 

             2Ag+ 2H₂SO₄→ Ag₂SO₄+ 2H₂O+ SO₂

(mol)  0,04                              0,02                                                       Zn+ H₂SO₄→ ZnSO₄+ H₂

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

6,25g một chất rắn không tan là: Ag₂SO₄

→ \(n_{Ag_2SO_4}=\dfrac{m}{M}=\dfrac{6,25}{312}=0,02\left(mol\right)\)

=> m_Ag=n.M=0,04.108=4,32(g)

      m_Zn=n.M= 0,25.65=16,25(g)

→ m_{hh kim loại}=m_Ag+m_Zn=4,32+16,25=20,57(g)

=> %m_Ag=\(\dfrac{4,32}{20,57}.100\%=21\%\)

     %m_Zn= 100%- 21%= 79%

         2Ag+2H₂SO₄→Ag₂SO₄+2H₂O+SO₂

(mol) 0,04                   0,02                                                          

           Zn+H₂SO₄→ZnSO₄+H₂

(mol) 0,25                              0,25                                               

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

6,25g một chất rắn không tan đó là: Ag₂SO₄

→ \(n_{Ag_2SO_4}=\dfrac{m}{M}=\dfrac{6,25}{312}=0,02\left(mol\right)\)

m_Ag=n.M=0,04.108=4,32(g)

m_Zn=n.M=0,25.65=16,25(g)

m_{hh kim loại}= m_Ag+ m_Zn=4,32+16,25= 20,57(g)

=> %m_Ag=\(\dfrac{4,32}{20,57}.100\%=21\%\)

     %m_Zn= 100%-21%= 79%

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{3,7185}{24,79}=0.,15(mol)\\ \Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{10}.100\%=97,5\%\\ \Rightarrow \%_{Cu}=100\%-97,5\%=2,5\%\\ b,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{14\%}=78,21(g)\)

\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1         2              1            1

          0,2                                     0,2

b) \(n_{Zn}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{Zn}=0,2.65=13\left(g\right)\)

\(m_{Cu}=19,4-13=6,4\left(g\right)\)

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