8 hóa học em cần nhanh trong tối nay

$\rm 2.$

$\rm a)$

$\rm n_{Al}=a(mol), n_{Fe}=b(mol)$

$\rm\rightarrow 27a+56b=0,83$

$\rm 2Al+3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2$

$\rm Fe+H_2SO_4 \rightarrow FeSO_4+H_2$

$\rm n_{H_2}=$  $\rm\dfrac{0,56}{22,4}=0,025(mol)$ 

$\rm\rightarrow n_{H_2}=$ $\rm\dfrac{3}{2}n_{Al}+n_{Fe}=$ $\rm\dfrac{3}{2}a+b=0,025(2)$ 

$\rm (1),(2) \rightarrow a=b=0,01$

$\rm\rightarrow \%Al=$ $\rm\dfrac{0,01×27}{0,83}×100\%\approx 32,53\%$ 

$\rm\rightarrow \%Fe=100\%-32,53\%=67,47\%$

$\rm b)$ $\rm n_{H_2SO_4}=n_{H_2}=0,025(mol)$

$\rm\rightarrow m_{H_2SO_4}=0,025×98=2,45(g)$

$\rm c)$ $\rm H_2+CuO \xrightarrow[]{t^o}H_2O+Cu$

$\rm n_{Cu}=n_{H_2}=0,025(mol)$

$\rm m_{\text{Chất rắn}}=m_{Cu}=0,025×64=1,6(g)$

`a)`

Đặt `n_{Al}=x(mol);n_{Fe}=y(mol)`

`->27x+56y=0,83(1)`

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

`Fe+H_2SO_4->FeSO_4+H_2`

Theo PT: `1,5x+y=n_{H_2}=0,025(2)`

`(1)(2)->x=y=0,01(mol)`

`\%m_{Al}={27.0,01}/{0,83}.100\%\approx 32,5\%`

`\%m_{Fe}=100-32,5=67,5\%`

`b)`

Theo PT: `n_{H_2SO_4}=n_{H_2}=0,025(mol)`

`->m_{H_2SO_4}=0,025.98=2,45(g)`

`c)`

`CuO+H_2`  $\xrightarrow{t^o}$  `Cu+H_2O`

`0,025←0,025->0,025(mol)`

`->m_{Cu}=0,025.64=1,6(g)`