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$a)PTHH:Fe+2HCl\to FeCl_2+H_2$
$\Rightarrow n_{Fe}=n_{H_2}=\dfrac{2,479}{24,79}=0,1(mol)$
$\Rightarrow \%m_{Fe}=\dfrac{0,1.56}{12}.100\%=46,67\%$
$\Rightarrow \%m_{Cu}=100-46,67=53,33\%$
$b)n_{FeCl_2}=n_{Fe}=0,1(mol)$
$\Rightarrow m_{FeCl_2}=0,1.127=12,7(g)$
$c)n_{HCl}=2n_{Fe}=0,2(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M$
\(a,n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,6(mol)\\ \Rightarrow m_{CT_{HCl}}=0,6.36,5=21,9(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{21,9}{28\%}=78,21(g)\\ b,n_{Mg}=n_{H_2}=0,3(mol)\\ \Rightarrow m_{Mg}=0,3.24=7,2(g)\\ \Rightarrow {\%}_{Mg}=\dfrac{7,2}{18}.100{\%}=40\%\\ \Rightarrow {\%}_{Ag}=60\%\)
a, Ta có: 27nAl + 56nFe = 22 (1)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)
b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ b,m_{hh}=5,4+12,8=18,2(g)\\ c,n_{HCl}=0,6(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M\)
a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,2<----0,6<---------------0,3
=> mAl = 0,2.27 = 5,4(g)
b) mhh = 5,4 + 12,8 = 18,2(g)
c) \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\)
a)
\(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
_____0,4<---0,4<--------0,4<----0,4
=> mZn = 0,4.65 = 26 (g)
=> \(\left\{{}\begin{matrix}\%Zn=\dfrac{26}{51,6}.100\%=50,388\%\\\%Cu=\dfrac{51,6-26}{51,6}.100\%=49,612\text{%}\end{matrix}\right.\)
b)
mZnSO4 = 0,4.161 = 64,4 (g)
c)
\(V_{ddH_2SO_4}=\dfrac{0,4}{2}=0,2\left(l\right)\)
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{3,7185}{24,79}=0.,15(mol)\\ \Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{10}.100\%=97,5\%\\ \Rightarrow \%_{Cu}=100\%-97,5\%=2,5\%\\ b,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{14\%}=78,21(g)\)
$n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)$
$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,2.27}{24,6}.100\%=21,95\%$
$\Rightarrow \%m_{Cu}=100-21,95=78,05\%$
$b)n_{AlCl_3}=n_{Al}=0,2(mol)$
$\Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)$
$c)n_{HCl}=3n_{Al}=0,6(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M$