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a)
Gọi $n_{Fe} = a(mol) ; n_{Al} =b (mol) \Rightarrow 56a + 27b = 11(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : $n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$
Từ (1)(2) suy ra : a = 0,1 ; b = 0,2
$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,9\%$
$\%m_{Al} = 100\% - 50,9\% = 49,1\%$
b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,8}{0,4} = 2M$
c)
$C_{M_{FeCl_2}} = \dfrac{0,1}{0,4} = 0,25M$
$C_{M_{AlCl_3}} =\dfrac{0,2}{0,4} = 0,5M$
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{Fe}=0,1(mol);n_{HCl}=0,2(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ C_{M_{HCl}}=\dfrac{0,2}{0,25}=0,8M\\ b,m_{dd_{HCl}}=250.1,12=280(g)\\ n_{FeCl_2}=0,1(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,1.127}{5,6+280-0,1.2}.100\%=4,45\%\)
\(a)Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ n_{Fe}=n_{H_2}=0,1mol\\ m_{Fe}=0,1.56=5,6g\\ m_{Fe_2O_3}=21,6-5,6=16g\\ c)n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\\ n_{H_2SO_4}=0,1+0,1.3=0,4mol\\ C_{M_{H_2SO_4}}=\dfrac{0,4}{0,5}=0,8M\)
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)
a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al}:x\left(mol\right)\\n_{Fe}:y\left(mol\right)\end{matrix}\right.\)
Ta có : \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\)
=> x=0,2 ; y=0,1
\(\%m_{Al}=\dfrac{0,2.27}{11}.100==49,09\%\)
\(\%m_{Fe}=50,91\%\)
b) \(\Sigma n_{HCl}=3x+2y=0,8\left(mol\right)\)
=> \(V_{HCl}=\dfrac{0,8}{2}=0,4\left(lít\right)\)
c) \(CM_{AlCl_3}=\dfrac{0,2}{0,4}=0,5M\)
\(CM_{FeCl_2}=\dfrac{0,1}{0,4}=0,25M\)
\(PTHH:4Al+6HCl\rightarrow2Al_2Cl_3+3H_2\uparrow\)
\(n_{Al}=\frac{3,78}{27}=0,14\left(mol\right)\)
\(\Rightarrow n_{H_2}=\frac{3}{4}n_{Al}=0,105\left(mol\right)\)
\(V_{H_2}=0,105.22,4=2,352\left(l\right)\)
\(n_{HCl}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,14=0,21\left(mol\right)\)
\(C_{M_{ddHCl}}=\frac{0,21}{0,2}=1,05\left(M\right)\)
\(n_{Al_2Cl_3}=\frac{1}{2}n_{Al}=\frac{1}{2}.0,14=0,07\left(mol\right)\)
\(m_{Al_2Cl_3}=0,07.160,5=11,235\left(g\right)\)
\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(n_{Fe_2O_3}=\dfrac{21,6-56.0,1}{160}=0,1mol\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2\)
0,1 0,6 0,2 0,3
\(V_{ddHCl}=\dfrac{0,2+0,6}{1}=0,8l\\ b.C_{M_{FeCl_2}}=\dfrac{0,1}{0,8}=0,125M\\ C_{M_{FeCl_3}}=\dfrac{0,2}{0,8}=0,25M\)