Đáp án:

$a) 68,4g$

$b) 13,44l$ 

Giải thích các bước giải:

Ta có PTHH sau:

$2Al+3H_2SO_4$ $\rightarrow$ $Al_2(SO_4)_3+3H_2$

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a)

Ta có $m_{H_2SO_4}=\dfrac{mdd.C}{100}=\dfrac{168.35}{100}=58,8g$

Vậy $n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{58,8}{98}=0,6$ mol

Dựa vào PTHH) $n_{Al_2(SO_4)_3}=\dfrac{0,6}{3}=0,2$ mol

Vậy $m_{Al_2(SO_4)_3}=n.M=0,2.342=68,4g$

b)

Dựa vào PTHH) $n_{H_2}=n_{H_2SO_4}=0,6$ mol

Vậy $V_{H_2}=0,6.22,4=13,44l$ 

`#Ly`

$a)$

`m_(H_2SO_4` `=` `(168.35)/(100)=58,8(g)`

`n_(H_2SO_4` `=` `(58,8)/(98)=0,6(mol)`

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

Theo PT: `n_(Al_2(SO_4)_3` `=` `(1)/(3)n_(H_2SO_4` `=` `0,2(mol)`

`m_(Al_2(SO_4)_3` `=` `0,2.342=68,4(g)`

$b)$

Theo PT: `n_(H_2` `=` `n_(H_2SO_4` `=` `0,6(mol)`

`V_(H_2` `=` `0,6.22,4=13,44(l)`