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\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow \%_{Fe}=\dfrac{5,6}{12}.100\%=46,67\%\\ \Rightarrow \%_{Cu}=100\%-46,67\%=53,33\%\\ c,n_{HCl}=2n_{H_2}=0,2(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a) Theo Pt : \(n_{H2}=n_{Fe}=n_{FeCl2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\)
\(\%m_{Cu}=100\%-56\%=44\%\)
b) Theo Pt : \(n_{H2}=2n_{HCl}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}.100\%=100\left(g\right)\)
c) \(m_{ddspu}=10+100-0,1.2=109,8\left(g\right)\)
\(C\%_{FeCl2}=\dfrac{0,1.127}{109,8}.100\%=11,57\%\)
\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2
b) \(n_{Fe}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(m_{Cu}=16-11,2=4,8\left(g\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
Chúc bạn học tốt
a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
____0,15<--0,3<--------------0,15
=> mFe = 0,15.56 = 8,4 (g)
=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{8,4}{21,2}.100\%=39,62\%\\\%Cu=\dfrac{21,2-8,4}{21,2}.100\%=60,38\%\end{matrix}\right.\)
b) mHCl = 0,3.36,5 = 10,95(g)
=> \(m_{ddHCl}=\dfrac{10,95.100}{3,65}.100\%=300\left(g\right)\)
a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,1 0,2 0,1
b,\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow\%m_{Fe}=\dfrac{5,6.100\%}{12}=46,67\%;\%m_{Cu}=100-46,67=53,33\%\)
c,\(m_{HCl}=0,2.36,5=7,3\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{7,3.100}{14,6}=50\left(g\right)\)
a) Fe tham gia phản ứng
$Fe + 2HCl \to FeCl_2 + H_2$
$m_{Fe} = 12 - 6,4 = 5,6(gam)$
b) $n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)$
Theo PTHH : $n_{H_2} = n_{Fe} = 0,1(mol)$
$\Rightarrow V_{H_2} = 0,1.22,4 = 2,24(lít)$