\(n_{H_2}=\dfrac{0,56}{22,4}=0,025(mol)\\ n_{Fe}=x(mol);n_{Al}=y(mol)\\ \Rightarrow 56x+27y=0,83(1)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow x+1,5y=0,025(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,01(mol)\\ y=0,01(mol) \end{cases}\Rightarrow \begin{cases} \%_{Fe}=\dfrac{0,01.56}{0,83}.100\%=67,47\%\\ \%_{Al}=100\%-67,47\%=32,53\% \end{cases}\)

a) Gọi `n_{Al} = a (mol); n_{Fe} = b (mol)`

PTHH:

`2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2`

`Fe + H_2SO_4 -> FeSO_4 + H_`

b) `n_{H_2} = (0,56)/(22,4) = 0,025 (mol)`

Theo PT: `n_{H_2} = n_{Fe} + 3/2 n_{Al}`

`=> b + 1,5a = 0,025`

Giải hpt \(\left\{{}\begin{matrix}27a+56b=0,83\\1,5a+b=0,025\end{matrix}\right.\Leftrightarrow a=b=0,01\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)

n_H2=0.56:22,4=0,025 mol 
Fe+H₂SO₄----->FeSO₄+H₂ 
2AL+3H2SO4----->AL2(SO4)3 +3H2 
Gọi x,y làn lượt là số mol Fe và AL 
ta có hệ pt 
$\{\begin{array}{c}56x+27y=0,83\\ x+1,5y=0,025\end{array}$
$\{\begin{array}{c}x=0,01mol\\ y=0,01mol\end{array}$
m_Fe=0,01.56=0,56 g 
m_Al=0,83-0,56=0,27 g 
%mFe=(0,56:0,83).100=67,47% 
%m_Al=100-67,47=32,53%

a. PTHH:

\(Zn+H_2SO_4--->ZnSO_4+H_2\)

\(Cu+H_2SO_4--\times-->\)

b. Theo PT: \(n_{Zn}=n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow m_{_{ }Zn}=0,1.65=6,5\left(g\right)\)

\(\Rightarrow\%_{m_{Zn}}=\dfrac{6,5}{10,5}.100\%=61,9\%\)

\(\%_{m_{Cu}}=100\%-61,9\%=31,8\%\)

\(Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\)

Cu không phản ứng H₂SO₄ loãng nhé

\(n_{H_2}= \dfrac{2,24}{22,4}= 0,1 mol\)

Theo PTHH:

\(n_{Zn}=n_{H_2}= 0,1 mol\)

\(\Rightarrow m_{Zn}= 0,1 . 65= 6,5 g\)

\(\Rightarrow\)%m_Zn=\(\dfrac{6,5}{10,5} . 100\)%~ 61,9%

\(\Rightarrow\)%m_Cu= 100% - 61,9%=38,1 %

\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b.n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{Zn}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Zn}=\dfrac{0,1.65}{10,5}.100=61,9\%\\ \%m_{Cu}=100-61,9=38,1\%\)

Gọi x, y lần lượt là số mol Al, Fe

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}27x+56y=0,83\\1,5x+y=0,025\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,01\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,27\left(g\right)\\m_{Fe}=0,56\left(g\right)\end{matrix}\right.\)

PTHH: 

Zn + H₂SO₄ ---> ZnSO₄ + H₂ (1)

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂ (2)

Ta có: \(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

Gọi x, y lần lượt là số mol của Zn và Al

a. Theo PT₍₁₎: \(n_{H_2}=n_{Zn}=x\left(mol\right)\)

Theo PT₍₂₎: \(n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}y\left(mol\right)\)

=> \(x+\dfrac{3}{2}y=0,8\) (*)

Theo đề, ta có: 65x + 27y = 3,79 (**)

Từ (*) và (**), ta có HPT:

\(\left\{{}\begin{matrix}x+\dfrac{3}{2}y=0,8\\65x+27y=3,79\end{matrix}\right.\)

(Ra số âm, bn xem lại đề nhé.)

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

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

b) Gọi x,y là số mol Al, Fe

\(n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

Ta có hệ : \(\left\{{}\begin{matrix}27x+56y=0,83\\\dfrac{3}{2}x+y=0,02\end{matrix}\right.\)

=> \(x=\dfrac{29}{5700};y=\dfrac{47}{3800}\)

\(\%m_{Al}=\dfrac{\dfrac{27}{5700}.27}{0,83}.100=16,55\%\); \(\%m_{Fe}=100-16,55=83,45\%\)

c)Bảo toàn nguyên tố H: \(n_{H_2SO_4}=n_{H_2}=0,02\left(mol\right)\)

=> \(C\%_{H_2SO_4}=\dfrac{0,02.98}{200}.100=0,98\%\)