Nhiệt phân hoàn toàn a g AL(OH)3 đến khồi lượng không đổi thu được 20,4 g chất rắn. Giá trị bằng số của a là

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Đáp án:mAl(OH)3=31,1 g

Giải thích các bước giải:

nAl2O3=m/M=20,4/102=0,2 (mol)

PTHH        2Al(OH)3→Al2O3+3H2O

                      0,4            0,2                mol

mAl(OH)3=n.M=0,4.78=31,2 (g)

 

`2Al(OH)_3→ Al_2O_3+ 3H_2O` 

`n_{Al_2O_3}= {20,4}/{102}= 0,2(mol)`

`n_{Al(OH)_3}= 2n_{Al_2O_3}= 0,4(mol)` 

`a= 0,4 .78= 31,2(g)`

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