Đáp án:

`↓` 

Giải thích các bước giải:

Ta có :

`n_{Al}=(4,5)/27=1/6(mol)`

`m_{ct_{H_2SO_4}}=(156,8xx15)/100=23,52(gam)`

`=>n_{H_2SO_4}=(23,52)/(98)=0,24(mol)`

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

`n_{Al}/2>n_{H_2SO_4}/3`

`1/12>0,08`

`=>H_2SO_4` hết và `Al` dư

`=>n_{H_2}=n_{H_2SO_4}=0,24(mol)`

`=>V_{H_2}=0,24xx22,4=5,376(lít)`

`b)`

`n_{Al_2(SO_4)_3}=1/3n_{H_2SO_4}=0,08(mol)`

`=>m_{Al_2(SO_4)_3}=0,08xx342=27,36(gam)`

`d)`

`m_{dd_{spứ}}=158,6+4,5-(0,24xx2)=162,62(gam)`

`=>C%_{Al_2(SO_4)_3}=(27,36xx100)/(162,62)=16,8(%)`

`#dkt`

$a,$

$m_{H_2SO_4}=\dfrac{156,8×15}{100}=23,52(g)$

$n_{H_2SO_4}=\dfrac{23,52}{98}=0,24(mol)$

$n_{Al}=\dfrac{4,5}{27}=\frac{1}{6}(mol)$

$2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2$

$0,16$          $0,24$            $0,08$              $0,24$

Xét $\dfrac{\dfrac{1}{6}}{2}>\dfrac{0,24}{3}$ $⇒\ Al$ dư

$→V=0,24×22,4=5,376(l)$

$b,$

$→m_{Al_2(SO_4)_3}=0,08×342=27,36(g)$

$c,$

$m_{dd}=0,16×27+156,8-0,25×2=160,62(g)$

$→C\%_{dd\ A}=\dfrac{27,36}{160,62}×100\%=17,03\%$