Đáp án:
25: 400ml
19: 0,35
21: 0,06
Giải thích các bước giải:
Câu 25:
Các phản ứng xảy ra:
$H^++OH^-\to H_2O\\Al^{3+}+3OH^-\to Al(OH)_3\\Al(OH)_3+OH^-\to AlO^-_2+2H_2O$
$n_{HCl}=0,5.0,2=0,1\ mol⇒n_{H^+}=0,1\ mol\\n_{Al_2(SO_4)_3}=0,25.0,2=0,05\ mol⇒n_{Al^{3+}}=0,1\ mol$
$a=n_{↓\ lớn\ nhất}=n_{Al(OH)_3}=n_{Al^{3+}}=0,1\ mol$
$n_{OH^-}=n_{H^+}+3.n_{Al^{3+}}=0,1+0,1.3=0,4\ mol⇒V=400ml$
Câu 19:
PTHH:
$H^++OH^-\to H_2O\\AlO_2+H^++H_2O\to Al(OH)_3\\Al(OH)_3+3H^+\to Al^{3+}+3H_2O$
⇒$n_{OH^-}=0,1\ mol⇒x=0,05$
$0,3=n_{OH^-}+4.n_{AlO^-_2}-3.n_↓⇒n_{AlO^-_2}=0,3 ⇒ y=0,15$
$x:y=1:3=0,33≈0,35$
Câu 21:
$Al^{3+}+3OH^-\to Al(OH)_3\\Al(OH)_3+OH^-\to AlO^-_2+2H_2O\\Ba^{2+}+SO^{2-}_4\to BaSO_4$
$69,9=m_{BaSO_4}⇒n_{SO^2-_4}=n_{BaSO_4}=0,3\ mol⇒a+3b=0,3$
$n_{Ba(OH)_2}=0,32⇒n_{OH^-}=0,64=4.n_{Al^{3+}}⇒n_{Al^{3+}}=0,16⇒b=0,08\\⇒a=0,06$
