a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

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

b) \(n_{H_2\left(1\right)}=n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{H_2\left(2\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.\dfrac{2,7}{27}=0,15\left(mol\right)\)

=> \(V=\left(0,2+0,15\right).22,4=7,84\left(l\right)\)

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

\(n_{H_2SO_4\left(2\right)}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{0,35.98}{20\%}=171,5\left(g\right)\)

d) \(m_{ddsaupu}=4,8+2,7+171,5-0,35.2=178,3\left(g\right)\)

\(C\%_{MgSO_4}=\dfrac{120.0,1}{178,3}.100=6,73\%\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,05}{178,3}.100=9,59\%\)

a,Mg+H2SO4-> MgSO4 +H2

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

b, n(Mg)=0,2mol

n(Al)=0,1mol

Số mol H2SO4=số mol H2= 0,2+ 0,1*3/2 =0,35mol

V(H2)= 7,84lit

c, MgSO4: m=0,2*120=24(g)

Al2(SO4)3 : m=342*0,05= 17,1(g)

d, khối lượng H2SO4= 0,35*98=34,3(g)

Khối lượng dd H2SO4 là: 

m(dd)=34,3*100/20 = 171,5(g)

e,khối lượng dd sau pứ 

m= m(Mg) +m(Al) + m(dd H2SO4) -m(H2) = 4,8+2,7+171,5-0,35*2=178,3(g)

C%(MgSO4)= 24*100%/178,3 =13,46%

C%(Al2SO4)3 = 17,1*100%/178,3 =9,59%

a)Mg+2CH3COOH→Mg(CH3COO)2+H2

Zn+2CH3COOH→Zn(CH3COO)2+H2

nH2=0,3mol

Gọi a và b lần lượt là số mol của Mg và Zn

\(\left\{{}\begin{matrix}24a+65b=11,3\\a+b=0,3\end{matrix}\right.\)

→a=0,2,b=0,1

→mMg=0,2×24=4,8g

→mZn=0,1×65=6,5g

b)%mMg=\(\dfrac{4,8}{11,3}\)×100%=42,48%

%mZn=\(\dfrac{6,5}{11,3}\)×100%=57,52%

c)nCH3COOH=2nMg+2nZn=0,6mol

→mCH3COOH=0,6×60=36g

→C%CH3COOH=\(\dfrac{36}{200}\)×100%=18%

→nMg(CH3COO)2=nMg=0,2mol

→nZn(CH3COO)2=nZn=0,1mol

→mMg(CH3COO)2=0,2×142=28,4g

→mZn(CH3COO)2=0,1×183=18,3g

nH2=nMg+nZn=0,3mol

→mH2=0,6g

→mddmuối=mhỗnhợp+mddCH3COOH−mH2

→mddmuối=11,3+200−0,6=210,7g

→C%Mg(CH3COO)2=\(\dfrac{28,4}{210,7}\)×100%=13,48%

→C%Zn(CH3COO)2=\(\dfrac{18,3}{210,7}\)×100%=8,69%

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ a,\%m_{Fe}=\dfrac{0,02.56}{4,36}.100\approx25,688\%\\ \Rightarrow\%m_{Ag}\approx74,312\%\\ b,Ta.thấy:2,18=\dfrac{1}{2}.4,36\\ \Rightarrow m_{hh\left(câuB\right)}=\dfrac{1}{2}.m_{hh\left(câuA\right)}\\ n_{Fe}=\dfrac{0,02}{2}=0,01\left(mol\right)\\ n_{Ag}=\dfrac{2,18-0,01.56}{108}=0,015\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ 2Ag+Cl_2\rightarrow\left(t^o\right)2AgCl\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+\dfrac{1}{2}.n_{Ag}=\dfrac{3}{2}.0,01+\dfrac{1}{2}.0,015=0,0225\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,0225.22,4=0,504\left(l\right)\)

Ta có: \(\left\{{}\begin{matrix}x=Fe\\y=Cu\end{matrix}\right.\) trong 40g hh

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

PTHH: Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

  TL:      1        2            1         1

mol:     0,5     \(\leftarrow\)  1    \(\leftarrow\)   0,5  \(\leftarrow\) 0,5

\(m_{Fe}=n.M=0,5.56=28g\)

\(\%m_{Fe}=\dfrac{m_{Fe}}{m_{hh}}.100\%=\dfrac{28}{40}.100\%=70\%\)

\(\%m_{Cu}=100\%-70\%=30\%\)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1                                   0,1   ( mol )

( Cu không tác dụng với dd axit HCl )

\(m_{Fe}=0,1.56=5,6g\)

\(\rightarrow m_{Cu}=12-5,6=6,4g\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{12}.100=46,66\%\\\%m_{Cu}=100\%-46,66\%=53,34\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\)

\(b,\) Đặt \(n_{Fe}=x(mol);n_{Al}=y(mol)\)

\(\Rightarrow 56x+27y=8,3(1)\)

Theo PTHH: \(x+1,5y=0,25(2)\)

\((1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8,3}.100\%=67,47\%\\ \%_{Al}=100\%-67,47\%=32,53\%\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Đặt:n_{Fe}=a\left(mol\right);n_{Zn}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}56a+65b=23,3\\a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Zn}=\dfrac{0,1.65}{23,3}.100\approx27,897\%\\ \Rightarrow\%m_{Fe}\approx72,103\%\)

Fe+2HCl--->FeCl2+H2
    Zn+2HCl-->ZnCl2+H2
Gọi số mol của Fe và Zn lần lượt là x,y mol
=> ta có hpt {56x+65y=23,3
                     {x+y=8,96/22,4
<=>{x=0,3=>mFe=16,8g
      {y=0,1=>mZn=6,5g
nHCl=2nH2=2.8,96/22,4=0,8 mol
=>mHCl=29,2g
%mFe=16,8/23,3.100=72,10300429%
=>%mZn=27,89699571%

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