Tìm kiếm
Nguyễn Gia Bảo
3 tuần trước

Hoà tan hết một oxit kim loại R ( có hoá trị ko đổi) bằng dd H2S04 39,2% vừa đủ, thu đc dd muooid có nồng nộ 40,14%. Tìm công thức hoá hc của oxit

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namthichhoa
3 tuần trước

Gọi `x` là hóa trị của kim loại :

Giả sử khối lượng oxit tham gia Phản ứng là 1(mol) là `1mol`
`R_2O_x + xH_2SO_4 -> R_2(SO_4)_x + xH_2O`

`m_{R_2O_x} = (2R+16x) (g)`
`m_{H_2SO_4} = 98x (g)`
`m_{dd_{H_2SO_4}} = (98x.100)/(39,2)=250x (g)`
`m_{dd}(spư)=2R+266x (g)`
`m_{R_2(SO_4)_x} = 2R+96x (g)`
Nồng độ muối sau phản ứng :
`(96x+2R)/(2R+266x)=(40,14)/100`
`<=> 119,72R=1077,24x`
`<=>R=9x`
Ta thấy `x=3`, `R=27` là thõa mãn 
Vậy `CT` oxit là `Al_2O_3`

`#namth`
 

hong_tham2007pbc
3 tuần trước

Đáp án:

 Cho `mdd _{H2SO4} = 100 g`

`-> nH2SO4 = 0,4`

`R2Ox + xH2SO4 -> R2(SO4)x + xH2O`

`(0,4)/x       0,4             (0,4)/x` 

`-> C%(R2(SO4)x) = ((0,8.R)/x + 38,4 )/[( 0,8.R)/x + 106,4 ] .100 = 40,14`

`-> R  = 27 (Al)`

-> `Al2O3`

mong hay nhất

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