lam giup minh vs a...

Tham khảo :

$n_{H_2}=0,02(mol)$

`-` Thêm $Mg$ vào còn phản ứng.

`=>`$Acid$ dư.

$n_{HCl}=0,3 \cdot 0,1=0,03(mol)$

$n_{H_2SO_4}=0,3 \cdot  0,2=0,06(mol)$

$n_{NaOH}=0,001V \cdot  0,2=0,0002V(mol)$

$n_{Ba(OH)_2}=0,001V \cdot  0,15=0,00015V(mol)$

$H_2SO_4+2NaOH \to Na_2SO_4+2H_2O$

$H_2SO_4+Ba(OH)_2 \to BaSO_4+2H_2O$

$HCl+NaOH \to NaCl+H_2O$

$2HCl+Ba(OH)_2 \to BaCl_2+2H_2O$

`-` Bảo toàn nguyên tố $H:$

$n_{HCl}+2n_{H_2SO_4}=2n_{H_2}+n_{NaOH}+2n_{Ba(OH)_2}$

`->`$0,03+0,06 \cdot 2=2 \cdot 0,02+0,0002V+2 \cdot 0,00015V$

`=>`$V=220(ml)$

$n_{Ba(OH)_2}=0,00015 \cdot 220=0,033(mol)$

$n_{H_2SO_4}=0,06(mol)$

$Ba(OH)_2+H_2SO_4 \to BaSO_4+2H_2O$

$n_{Ba(OH)_2}<n_{H_2SO_4}$`->`$H_2SO_4$ dư.

`=>`$n_{BaSO_4}=0,033(mol)$

`=>`$m=0,033 \cdot 233=7,689(gam)$

Do `Mg` vào `X` thoát `H_2` nên `X` chứa `A` dư

`n_{H_2}={0,448}/{22,4}=0,02(mol)`

`n_{H^+}=0,1.0,3+0,2.2.0,3=0,15(mol)`

`n_{SO_4^{2-}}=0,2.0,3=0,06(mol)`

`n_{Ba^{2+}}=0,001V.0,15=0,00015V(mol)`

`n_{OH^-}=0,2.0,001V+2.0,15.0,001V=0,0005V(mol)`

`H^{+}+OH^{-}->H_2O`

`0,0005V←0,0005V(mol)`

`Mg+2H^{+}->Mg^{2+}+H_2`

`0,02←0,04←0,02←0,02(mol)`

`->0,04+0,0005V=0,15`

`->V=220(ml)`

`->n_{Ba^{2+}}=0,00015.220=0,033(mol)`

`Ba^{2+}+SO_4^{2-}->BaSO_4`

`0,033->0,033->0,033(mol)`

`->m_↓=0,033.233=7,689(g)`