Đáp án+Giải thích các bước giải:

`2Al+3H_{2}SO_{4}→Al_{2}(SO_{4})_{3}+3H_{2}`

`a.n_{H_{2}SO_{4}}=0,2.0,2=0,04(mol)`

`PT→n_{Al}=2/3.n_{H_{2}SO_{4}}=(2)/(3).0,04=(0,08)/3(mol)`

`→m_{Al}=a=(0,08)/(3).27=0,72(g)`

`b.`

`C_{1}:`

`PT→n_{Al_{2}(SO_{4})_{3}}=1/3.n_{H_{2}SO_{4}}=(1)/(3).0,04=(0,04)/3(mol)`

`→m_{Al_{2}(SO_{4})_{3}}=(0,04)/(3).342=4,56(g)`

`C_{2}:`

`PT→n_{H_{2}}=n_{H_{2}SO_{4}}=0,04(mol)`

`→m_{Al_{2}(SO_{4})_{3}}=0,72+98.0,04-2.0,04=4,56(g)`

`n_{H_2SO_4}=0,2.0,2=0,04(mol)`

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

`a.n_{Al}=2/3 n_{H_2SO_4}=2/3 .0,04=2/75 (mol)`

`a=m_{Al}=2/75 .27=0,72(g)`

`b.`

`@` cách 1:

`n_{Al_2(SO_4)_3}=1/3 n_{H_2SO_4}=1/3 .0,04=1/75(mol)`

`=>m_{Al_2(SO_4)_3}=1/75 .342=4,56(g)`

`@` cách 2:

`n_{H_2}=n_{H_2SO_4}=0,04(mol)`

`m_{Al}+m_{H_2SO_4}=m_{Al_2(SO_4)_3}+m_{H_2}`

`=>m_{Al_2(SO_4)_3}=0,72+0,04.98-0,04.2=4,56(g)`