Đáp án:

 `↓`

Giải thích các bước giải:

 `n_{Al}=(0,54)/(27)=0,02(mol)`

`m_{H_2SO_4}=40.9,8\%=3,92(g)`

`->n_{H_2SO_4}=(3,92)/(98)=0,04(mol)`

PTHH:

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

Lập tỉ lệ :

`(0,02)/2 < (0,04)/3`

`->` Sau phản ứng `H_2SO_4` dư , `Al` hết

`->n_{H_2}=3/2 n_{H_2SO_4}=0,03(mol)`

`->V=0,03.22,4=0,672(lít)`

`n_{Al_2(SO_4)_3}= 1/2 n_{Al}= 1/2 . 0,02=0,01(mol)`

`n_{H_2SO_4 dư}=0,04-0,02. 3/2 =0,01(mol)`

$m_{\text{dung dịch sau pứ}}=0,54+40-0,03.2=40,48(g)$

`C\%_{Al_2(SO_4)_3}=( 0,01.342)/(40,48).100\%≈8,45\%`

`C\%_{H_2SO_4 dư}=( 0,01.98)/(40,48).100\%≈2,42\%`

Đáp án + Giải thích các bước giải:

$n_{Al} = \dfrac{m}{M} = \dfrac{0,54}{27} = 0,02\text{ mol}$
$m_{H_2SO_4} = \dfrac{m_{dd} . C\%}{100\%} = \dfrac{40 . 9,8\%}{100\%} = 3,92g$

$n_{H_2SO_4} = \dfrac{m}{M} = \dfrac{3,92}{2 . 1 + 32 + 4 . 16} = \dfrac{3,92}{98} = 0,04\text{ mol}$

 PTHH:
$2Al + 3H_2SO_4 \longrightarrow Al_2(SO_4)_3 + 3H_2 \uparrow$

Ta có:

$\dfrac{n_{H_2SO_4}}{3} > \dfrac{n_{Al}}{2} \bigg(\dfrac{0,4}{3} > \dfrac{0,2}{2}\bigg)$

$\Rightarrow H_2SO_4$ dư, tính theo $Al$

Theo PTHH:
$n_{H_2} = \dfrac{3}{2}n_{Al} = \dfrac{3}{2} . 0,02 = 0,03\text{ mol}$
$\Rightarrow V_{H_2} = n . 2,24 = 0,03 . 22,4 = 0,672l$

$m_{H_2} = n . M = 0,03 . 2 = 0,06g$

$m_{ddAl_2(SO_4)_3} = m_{Al} + m_{ddH_2SO_4} - m_{H_2} - 0,54 + 40 - 0,06 = 40,48g$

Theo PTHH:

$n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = \dfrac{1}{2} . 0,02 = 0,01\text{ mol}$

$\Rightarrow m_{Al_2(SO_4)_3} = n . M = 0,01 . [2 . 27 + 3 . (32 + 4 . 16)] = 3,42g$

$C\%_{Al_2(SO_4)_3} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{3,42}{40,48} . 100\% \approx 8,45\%$ 

$n_{H_2SO_4\text{ dư }} = 0,04 - 0,03 = 0,01\text{ mol}$

$\Rightarrow m_{H_2SO_4\text{ dư }} = n . M = 0,01 . 98 = 0,98g$

$\Rightarrow C\%_{H_2SO_4\text{ dư }} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{0,98}{40,48} . 100\% \approx 2,42\%$