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m dd sau pư = mFe + m dd HCl - mH2 thôi em nhé, Cu không phản ứng nên không cộng thêm vào.
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
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) \(m_{ddspu}=10+100-0,1.2=109,8\left(g\right)\)
\(C\%_{FeCl2}=\dfrac{0,1.127}{109,8}.100\%=11,57\%\)
Vì Cu không tác dụng với HCl, nên chỉ có phản ứng của Fe.
PTHH: Fe+2HCl\(\rightarrow\)FeCl2+H2
a) nH2=0,15(mol)
Theo pt: nFe=nH2=0,15 (mol)
\(\Rightarrow\)mFe=8,4(g)
b) mCu=10-8,4=1,6(g)
c) Theo pt: nHCl=nH2=0,15(mol)
\(\Rightarrow\)VHCl=0,3(l)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\) (1)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (2)
a) Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=n_{Mg}\) \(\Rightarrow m_{Mg}=1\cdot24=24\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{24}{32}\cdot100\%=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)
b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\dfrac{32-24}{40}=0,4mol\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{HCl}=2,4mol\) \(\Rightarrow m_{ddHCl}=\dfrac{2,4\cdot36,5}{7,3\%}=1200\left(g\right)\)
c) Theo PTHH: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=1,2mol\)
\(\Rightarrow\Sigma m_{MgCl_2}=1,2\cdot95=114\left(g\right)\)
Mặt khác: \(m_{H_2}=1\cdot2=2\left(g\right)\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,15<-0,3<----0,15<---0,15
\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)
c) mHCl = 0,3.36,5 = 10,95 (g)
=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)
d) mdd = 12 + 109,5 - 0,15.2 = 121,2 (g)
\(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)
a) Gọi $n_{Al} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 33,4(1)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : $n_{H_2} = 1,5a + b = \dfrac{17,92}{22,4} = 0,8(2)$
Từ (1)(2) suy ra : a = 0,2 ; b = 0,5
$\%m_{Al} = \dfrac{0,2.27}{33,4}.100\% = 16,17\%$
$\%m_{Fe} = 100\% - 16,17\% = 83,83\%$
b) $n_{HCl} = 2n_{H_2} = 1,6(mol)$
c) $m_{muối} = m_{hh} + m_{HCl} - m_{H_2} = 33,4 + 1,6.36,5 - 0,8.2 = 90,2(gam)$
\(a/n_{khí}=\dfrac{2,24}{22,4}=0,1mol\\ n_{Fe}=a;n_{FeCO_3}=b\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ FeCO_3+2HCl\rightarrow FeCl_2+CO_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}56a+116b=7,2\\a+b=0,1\end{matrix}\right.\\ \Rightarrow a=\dfrac{11}{150};b=\dfrac{2}{75}\\ \%m_{Fe}=\dfrac{11:150.56}{7,2}\cdot100\%=57,04\%\\ \%m_{FeCO_3}=100\%-57,04\%=42,96\%\\ b/n_{FeCl_2}=\dfrac{11}{150}+\dfrac{2}{75}=0,1mol\\ C_{\%FeCl_2}=\dfrac{0,1.127}{7,2+94,9-\dfrac{11}{150}\cdot2-\dfrac{2}{75}\cdot44}\cdot100\%=12,6\%\)