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![](https://rs.olm.vn/images/avt/0.png?1311)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{Zn}=n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \%m_{Zn}=\dfrac{0,1.65}{10}.100=65\%\\ \Rightarrow\%m_{Cu}=100\%-65\%=35\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: Zn + H2SO4 -> ZnSO4+ H2
nH2= 0,1(mol) -> nZn=nH2=0,1(mol)
=> mZn=0,1.65=6,5(g)
=> %mZn=(6,5/10).100=65%
=> %mCu=100% - 65%= 35%
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ m_{Fe}=0,1.56=5,6\left(g\right)\\ \%m_{Fe}=\dfrac{5,6}{8}.100=70\%\\ \Rightarrow\%m_{Cu}=100\%-70\%=30\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
nH2 = \(\dfrac{2,24}{22,4}\)= 0,1 (mol)
PTHH: Zn + H2SO4 ----> ZnSO4 + H2\(\uparrow\)
______0,1 mol<----------------------0,1 mol
Cu không tác dụng với H2SO4
mZn = 0,1 .65 = 6,5 (g)
mCu = 10 - 6,5 = 3,5 (g)
%Zn = \(\dfrac{6,5}{10}\) . 100% = 65%
%Cu = 100 - 65 = 35%
Bài 2:
nH2 = \(\dfrac{5,6}{22,4}\) = 0,25 (mol)
Sau pứ thấy còn 6,25g chất rắn không tan là Ag
PTHH:
Zn + H2SO4 ----> ZnSO4 + H2\(\uparrow\)
0,25 mol<------------------0,25 mol
mZn = 0,25 . 65 = 16,25(g)
mhh = 6,25 + 16,25 = 22,5 (g)
%Ag = \(\dfrac{6,25}{22,5}\) . 100% = 27,78%
%Zn = 100 - 27,78 = 72,22%
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,075<-------------------------0,075
Cu không phản ứng với H2SO4 loãng
b, \(m_{Mg}=0,075.24=1,8\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,8}{8}.100\%=22,5\%\\\%m_{Cu}=100\%-22,5\%=77,5\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,075 0,075
\(Cu+H_2SO_4-x->\)
\(m_{Mg}=0,075.24=1,6\left(g\right)\\ m_{Cu}=8-1,6=6,4\left(g\right)\)
\(\%m_{Cu}=\dfrac{6,4}{8}.100\%=80\%\\
\%m_{Mg}=100-80\%=20\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A.Mg + H2SO4 --> MgSO4 + H2
Cu + H2SO4 -×->(không pư)
B. nH2 = 2,24/22,4 = 0,1(mol)
nMg = nH2 = 0,1mol
mMg = 0,1.24 = 2,4 (g)
mCu = 10 - 2,4 = 7,6(g)
C. %Mg = 2,4/10 ×100 = 24%
%Cu = 100 - 24 = 76%
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,05 0,05 ( mol )
( Cu không tác dụng với dd axit H2SO4 loãng )
\(m_{Mg}=0,05.24=1,2g\)
\(\rightarrow m_{Cu}=8-1,2=6,8g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,2}{8}.200=15\%\\\%m_{Cu}=100\%-15\%=85\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,1 ( mol )
( Cu không tác dụng với dd axit HCl )
\(m_{Fe}=0,1.56=5,6g\)
\(\rightarrow m_{Cu}=12-5,6=6,4g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{12}.100=46,66\%\\\%m_{Cu}=100\%-46,66\%=53,34\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi $n_{Fe} = a(mol), n_{Zn} = b(mol) , n_{Al} = c(mol) \Rightarrow 56a + 65b + 27c = 20,4(1)$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$2Al +3 H_2SO_4 \to Al_2(SO_4)_3 +3 H_2$
Theo PTHH : $n_{H_2} = a + b + 1,5c = \dfrac{10,08}{22,4} = 0,45(mol)(2)$
Mặt khác :
$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$Zn + Cl_2 \xrightarrow{t^o} ZnCl_2$
$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
Theo PTHH : $n_{Cl_2} = 1,5n_{Fe} + n_{Zn} + 1,5n_{Al}$
Suy ra : \dfrac{1,5a + b + 1,5c}{a + b + c} = \dfrac{0,275}{0,2}(3)$
Từ (1)(2)(3) suy ra : a = 0,2 ; b = 0,1 ; c = 0,1
$\%m_{Fe} = \dfrac{0,2.56}{20,4}.100\% = 54,9\%$
$\%m_{Zn} = \dfrac{0,1.65}{20,4}.100\% = 31,9\%$
$\%m_{Al} = 100\% - 54,9\% - 31,9\% = 13,2\%$
n H2=2.24/22.4=0.1(mol)
ZnSO4 + H2![phimtren](https://gianghi.net/forum/styles/default/xenforo/smilies/tren.png)
Zn + H2SO4
0.1 0.1
m Zn=0.1*65=6.5(g)
%Zn=6.5/10*100%=65%
%Cu=100%-65%=35%
Zn + H2SO4 -> ZnSO4 + H2 (1)
nH2=0,1(mol)
Từ 1:
nZn=nH2=0,1(mol)
mZn=65.0,1=6,5(g)
%mZn=\(\dfrac{6,5}{10}.100\%=65\%\)
%mCu=100-65=35%