Hấp thụ hoàn toàn 0.15mol so2 vào 400ml dd naoh x mol /l thứ được 16.7 gam muối. Giá trị x là

Đáp án:

$x = 0,5$

Giải thích các bước giải:

 * Nếu chỉ tạo muối $NaHSO_3$:

PTHH: 

$\begin{gathered} S{O_2} + NaOH \to NaHS{O_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,15 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \end{gathered} $

$\to{m_{muối}} = 104.0,15 = 15,6\,\,gam$

* Nếu chỉ tạo muối $Na_2SO_3$

PTHH:

$\begin{gathered} \,\,\,\, S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,15 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \to {m_{muối}} = 0,15.126 = 18,9\;\,gam\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} $

* Nhận thấy: $15,6 < 16,7 < 18,9 \to $ phản ứng tạo hai muối $NaHS{O_3}\,\,và\,\,N{a_2}S{O_3}$

Gọi số mol của $NaHS{O_3}\,\,và\,\,N{a_2}S{O_3}$ lần lượt là a và b mol.

PTHH:

$\begin{gathered} S{O_2} + NaOH \to NaHS{O_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ← \,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\,\,\,\,\,\,\,\,\,\, \hfill \\ b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ← \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \end{gathered} $

$\left\{ \begin{gathered} a + b = 0,15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 104a + 126b = 16,7 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} a = 0,1 \hfill \\ b = 0,05 \hfill \\ \end{gathered} \right.$

Số mol của NaOH là: ${n_{NaOH}} = a + 2b = 0,1 + 0,05.2 = 0,2\,\,mol$

${C_M} = \dfrac{n}{V} = \dfrac{{0,2}}{{0,4}} = 0,5M = x$