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PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\\%m_{Fe}=\dfrac{0,1\cdot56}{12}\cdot100\%\approx46,67\%\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\\%m_{Cu}=53,33\%\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Cu không phản ứng
\(nH_2=nFe=\dfrac{2,24}{22,4}=0,1mol\)
\(\rightarrow mFe=0,1.56=5,6gam\)
\(\rightarrow\%mFe=\dfrac{5,6}{12}.100\%=46,\left(6\right)\%\)
\(\rightarrow\%mCu=100\%-46,\left(6\right)\%=53,\left(3\right)\%\)
c)
\(CM_{HCl}=\dfrac{0,1.2}{0,2}=1M\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow \%_{Fe}=\dfrac{5,6}{12}.100\%=46,67\%\\ \Rightarrow \%_{Cu}=100\%-46,67\%=53,33\%\\ c,n_{HCl}=2n_{H_2}=0,2(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M\)
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
nHCl = 0,6+0,4 = 1 (mol)
\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)\)
Gọi a, b lần lượt là mol của Al và Zn
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a 1,5a
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b b
\(\Rightarrow\left\{{}\begin{matrix}27a+65b=9,2\\1,5a+b=\dfrac{5,6}{22,4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,1.27}{9,2}.100\%=29,35\%\)
\(\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%=70,35\%\)
b. \(n_{H_2}=0,25mol\) \(\Rightarrow n_{HCl}=0,5mol\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25g\)
Ta có: \(10\%=\dfrac{18,25}{m_{dd}}.100\%\)
\(\Leftrightarrow m_{dd}=182,5g\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,2(mol)\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow m_{Cu}=20-5,6=14,4(g)\\ c,\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\ \%m_{Cu}=100\%-28\%=72\%\)