a) $Fe + 2HCl \to FeCl_2 + H_2$

b) Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$\%m_{Fe} = \dfrac{0,2.56}{18}.100\% = 62,22\%$

$\%m_{Cu} = 100\% - 62,22\% = 37,78\%$

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,4.56}{35}.100\%=64\%\\\%m_{Cu}=36\%\end{matrix}\right.\)

giúp mình với =((

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(\Rightarrow n_{Al}=\dfrac{2}{3}\cdot n_{H_2}=\dfrac{2}{3}\cdot0.15=0.1\left(mol\right)\)

\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

Cu không phản ứng với HCl 

\(m_{Cu}=m_{hh}-m_{Al}=9.1-2.7=6.4\left(g\right)\)

\(\%m_{Al}=\dfrac{2.7}{9.1}\cdot100\%=29.67\%\)

\(\%m_{Cu}=100\%-29.67\%=70.33\%\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\)

\(b,\) Đặt \(n_{Fe}=x(mol);n_{Al}=y(mol)\)

\(\Rightarrow 56x+27y=8,3(1)\)

Theo PTHH: \(x+1,5y=0,25(2)\)

\((1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8,3}.100\%=67,47\%\\ \%_{Al}=100\%-67,47\%=32,53\%\)

a) $Fe + 2HCl \to FeCl_2 + H_2$

b) Theo PTHH : 

$n_{Fe} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$\%m_{Fe} = \dfrac{0,1.56}{16}.100\% = 35\%$

$\%m_{Cu} = 100\% - 35\% = 65\%$

tham khảo:

a) Fe + 2HCl --> FeCl2 + H2
    0,3                                0,3
b) n khí = nH2 = 6,721/22,4 = 0,3 (mol)
mFe = 0,3.56 = 16,8g
% Fe = 16,8.100/30 = 56%
%Cu = 100 - 56 = 44%

Em coi lại đề xem đúng chưa, chứ anh thấy cái thể tích khí số xấu lắm

nH2=4,968/22,4~0,22(mol)

A),PTHH:

Fe+2HCl→FeCl2+H2

B)nFe=nH2~0,22(mol)

⇒mFe=0,22.56~12,42(g)

⇒%Fe= 12,42 /20 .100 % = 62,1 %

⇒%Cu=100%−62,1%=37,9%

C)

nHCl=2nH2=0,44(mol)

⇒CnHCl=0,44/0,44=1M

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow \%_{Fe}=\dfrac{5,6}{12}.100\%=46,67\%\\ \Rightarrow \%_{Cu}=100\%-46,67\%=53,33\%\\ c,n_{HCl}=2n_{H_2}=0,2(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M\)