Cho hh gồm `0,3` mol `Ba(OH)_2` và `0,36`mol `KOH` vào dung dịch chứa : `Al^(3+)` (0,15 mol) ; `Mg^(2+)` (0,15 mol) ; `Cl^-` (0,2 mol) ; `SO_4^(2-)` (0,3 mol) ; `H^+` (0,05 mol) . Xác định sản phẩm thu đc sau phản ứng , có giải thích rõ ràng.

1 câu trả lời

Đáp án:

\({H_2}O,Al{(OH)_3},Mg{(OH)_2},BaS{O_4}\)

Giải thích các bước giải:

\(\begin{array}{l}
Ba{(OH)_2} \to B{a^{2 + }} + 2{\rm{O}}{H^ - }\\
K{\rm{O}}H \to {K^ + } + O{H^ - }\\
{n_{O{H^ - }}} = 2{n_{Ba{{(OH)}_2}}} + {n_{K{\rm{O}}H}}\\
 \to {n_{O{H^ - }}} = 2 \times 0,3 + 0,36 = 0,96mol\\
{H^ + } + O{H^ - } \to {H_2}O\\
A{l^{3 + }} + 3{\rm{O}}{H^ - } \to Al{(OH)_3}\\
M{g^{2 + }} + 2{\rm{O}}{H^ - } \to Mg{(OH)_2}\\
B{a^{2 + }} + S{O_4}^{2 - } \to BaS{O_4}\\
{n_{O{H^ - }}} = {n_{{H^ + }}} + 3{n_{A{l^{3 + }}}} + 2{n_{M{g^{2 + }}}} = 0,8mol < 0,96mol
\end{array}\)

Suy ra ion \(O{H^ - }\) dư 

Nên \(Al{(OH)_3}\) sẽ bị hòa tan hết 

\(Al{(OH)_3} + O{H^ - } \to Al{O_2}^ -  + 2{H_2}O\)

Vậy sản phẩm thu được sau phản ứng gồm: \({H_2}O,Al{O_2}^ -,Mg{(OH)_2},BaS{O_4}\)

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