Mọi người giúp

`n_{H_2}={5,04}/{22,4}=0,225(mol)`

`n_{Al_2(SO_4)_3}={273,75.21,863\%}/{342}\approx 0,175(mol)`

Bảo toàn S: `n_{SO_4^{2-}}=3n_{Al_2(SO_4)_3}=0,525(mol)`

`->m_{dd\ H_2SO_4}={0,525.98}/{20\%}=257,25(g)`

BTKL: `m=273,75+0,225.2-257,25=16,95(g)`

Đáp án:

`m=20,325(g)`

Giải thích các bước giải:

`2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2`

`2Al(OH)_3+3H_2SO_4 \to Al_2(SO_4)_3+3H_2O`

`Al_2O_3+3H_2SO_4 \to Al_2(SO_4)_3+3H_2O`

`n_{Al}=2/3n_{H_2}=2/3 . (5,04)/(22,4)=0,225(mol)`

`-` Theo pthh

`n_{Al_2(SO_4)_3}=(273,75.21,863%)/(342)=0,175(mol)`

`n_{Al_2(SO_4)_3(tao\ H_2O)}=0,175-(0,225)/2=0,0625(mol)`

`n_{H_2O}=3n_{Al_2(SO_4)_3(tao\ H_2O)}=3.0,0625=0,1875(mol)`

`n_{H_2SO_4}=3/2n_{Al}+n_{H_2O}=3/2 . 0,225+0,1875=0,525(mol)`

`=>m_{dd\ H_2SO_4}=(0,525.98.100)/(20)=257,25(g)`

`=>m+257,25=273,75+0,225.2`

`=>m=16,95`