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Gọi \(n_{Fe}=x\left(mol\right)\)\(;n_{Zn}=y\left(mol\right)\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
Ta có: \(\left\{{}\begin{matrix}56x+65y=18,6\\2x+2y=2n_{H_2}=0,6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,1\cdot56}{18,6}\cdot100\%=30,11\%\)
\(\%m_{Zn}=100\%-30,11\%=69,89\%\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4
\(n_{HCl}=0,2+0,4=0,6mol\)
\(C_M=\dfrac{n}{V}=\dfrac{0,6}{0,2}=3M\)
\(n_k=n_{H_2}=0,125\left(mol\right)\)
a,b, \(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
.............0,125...0,125....................0,125...
\(\Rightarrow m_{Fe}=7\left(g\right)\)
Do Cu không phản ứng với H2SO4 .
\(\Rightarrow m_{Cu}=m_{hh}-m_{Fe}=10-7=3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Fe=70\%\\\%Cu=30\%\end{matrix}\right.\)
c, Có : \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=206,75\left(g\right)\)
\(\Rightarrow C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%\approx5,925\%\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
_____0,15<--------------0,15<---0,15
=> mFe = 0,15.56 = 8,4 (g)
=> mCu = 11,6 - 8,4 = 3,2 (g)
\(\left\{{}\begin{matrix}\%Fe=\dfrac{8,4}{11,6}.100\%=72,414\%\\\%Cu=\dfrac{3,2}{11,6}.100\%=27,586\%\end{matrix}\right.\)
mFeSO4 = 0,15.152 = 22,8 (g)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,3\left(mol\right)=n_{ZnCl_2}\\n_{HCl}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0.3\cdot65}{35,5}\cdot100\%\approx54,93\%\\\%m_{Cu}=45,07\%\\C\%_{HCl}=\dfrac{0,6\cdot36,5}{500}\cdot100\%=4,38\%\\m_{ZnCl_2}=0,3\cdot136=40,8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{Cu}=35,5-0,3\cdot65=16\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{Cu}-m_{H_2}=518,9\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{518,9}\cdot100\%\approx7,86\%\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=0,1(mol)\\ a,\%_{Zn}=\dfrac{0,1.65}{9,7}.100\%=67,01\%\\ \Rightarrow \%_{Cu}=100\%-67,01\%=32,99\%\\ b,n_{HCl}=2n_{H_2}=0,2(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,2.36,5}{50}.100\%=14,6\%\\ c,PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{H_2SO_4}=n_{H_2}=0,1(mol)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5(l)\)
nH2 = 3,36/22,4 = 0,15 (mol)
Chỉ có Zn phản ứng với H2SO4
Zn + H2SO4 \(\rightarrow\) ZnSO4 + H2 ( tính theo H2 )
0,15 : 0,15 : 0,15 : 0,15
1) mZn = 0,15 . 65 = 9,75 (g)
mCu = 15 - 9,75 = 5,25 (g)
2) % mZn = 9,75 : 15 . 100% = 65%
% mCu = 100 - 65 = 35%
3) 100 ml = 0,1 l
CMH2SO4 = 0,15 / 0,1 = 1,5 M
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