\(a,PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b,\text{Chất rắn ko tan là }Cu\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \text{Theo PT: }n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\\ \Rightarrow m=m_{Cu}=9,1-2,7=6,4\left(g\right)\\ c,\%_{Al}=\dfrac{2,7}{9,1}\cdot100\%\approx29,67\%\\ \%_{Cu}\approx100\%-29,67\%=70,33\%\\ d,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)

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 =((

\(a,n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

           0,075<-------------------------0,075

Cu không phản ứng với H₂SO₄ loãng

b, \(m_{Mg}=0,075.24=1,8\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,8}{8}.100\%=22,5\%\\\%m_{Cu}=100\%-22,5\%=77,5\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\) 
            0,075                                0,075 
         \(Cu+H_2SO_4-x->\) 
          \(m_{Mg}=0,075.24=1,6\left(g\right)\\ m_{Cu}=8-1,6=6,4\left(g\right)\) 
\(\%m_{Cu}=\dfrac{6,4}{8}.100\%=80\%\\ \%m_{Mg}=100-80\%=20\%\)

a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Giả sử: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow27x+56y=16,6\left(1\right)\)

Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(\Sigma n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y\left(mol\right)\)

\(\Rightarrow\dfrac{3}{2}x+y=0,05\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\\y=\end{matrix}\right.\)

Tới đây ra số mol âm, bạn xem lại đề nhé!

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

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

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

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(0.2.....................................0.2\)

\(m_{Fe}=0.2\cdot56=11.2\left(g\right)\) 

\(m_{Fe}=m_{hh}=11.2\left(g\right)\)

=> Đề sai