Question

[Phương-g Hà]

Trộn V₁l dd HCl 0,6M vs V₂l dd NaOH 0,4M thì thu đc 0,6l dd A . Biết PƯ xảy ra theo sơ đồ :

HCl + NaOH -> NaCl + H₂O

Tính V₁ và V₂ biết rằng 0,6l dd A có hể hòa tan hết 1,02g Al₂O₃

Answer 1

[Lặng Lẽ]

\(n_{HCl}=0,6v_1\left(mol\right)\)

\(n_{NaOH}=0,4v_2\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{1,02}{102}=0,01\left(mol\right)\)

\(HCl+NaOH-->NaCl+H_2O\)

TH1: HCl dư

\(HCl+NaOH-->NaCl+H_2O\)

n pu 0,4v₂............0,4v₂

n spu 0,6v₁-0,4v₂.........0

dd A: NaCl(tt), HCl dư

\(Al_2O_3+6HCl-->2AlCl_3+3H_2O\)

0,01.........0,06

\(\left\{{}\begin{matrix}0,6v_1-0,4_2=0,06\\v_1+v_2=0,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}v_1=0,3\left(l\right)\\v_2=0,3\left(l\right)\end{matrix}\right.\)

TH2: NaOH dư

\(HCl+NaOH-->NaCl+H_2O\)

n pu 0,6v₁.......0,6v₁

n spu 0.............0,4v₂-0,6v₁

dd A: NaCl(tt), NaOH dư

\(2NaOH+Al_2O_3-->2NaAlO_2+H_2O\)

0,02.........0,01

\(\left\{{}\begin{matrix}0,4v_2-0,6v_1=0,02\\v_1+v_2=0,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}v_2=0,38\left(l\right)\\v_1=0,32\left(l\right)\end{matrix}\right.\)