Question

  

Answer 1

Câu 7: 

Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Al_2O_3}=b\left(mol\right)\end{matrix}\right.\Rightarrow80a+102b=35,3\left(1\right)\)

PTHH: 

`CuO + H_2SO_4 -> CuSO_4 + H_2O`

`Al_2O_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2O`

Theo PT:

`n_{CuSO_4} = n_{CuO} = a (mol)`

`n_{Al_2(SO_4)_3} = n_{Al_2O_3} = b (mol)`

`=> 160a + 342b = 91,3 (2)`

`(1), (2) => a = 0,25; b = 0,15`

`=>` \(\%m_{Al_2O_3}=\dfrac{0,15.102}{35,3}.100\%=43,34\%\)

Câu 8: 

Gọi CTHH của oxit là \(R_xO_y\)

Ta có: \(\dfrac{m_R}{m_O}=\dfrac{\%m_R}{\%m_O}\)

`=>` \(\dfrac{xM_R}{16y}=2,5\Leftrightarrow M_R=\dfrac{40y}{x}=20.\dfrac{2y}{x}\left(g/mol\right)\)

Với `(2y)/x = 2 => M_R = 20.2 = 40 (g//mol)`

`=> R: Canxi(Ca)`

`X: CaO`

Answer 2

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