Hòa tan hết 4,667 gam hỗn hợp Na, K, Ba và ZnO vào nước

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} \begin{array}{l}+ \left\{\begin{array}{l}{n}_{ZnO}=n_O=\frac{4,667.5,14\%}{16}=0,015\\ n_{~nt (+)}=n_{O{H}^{-}}=n_{H^{+}/H_{2}O}=2n_{H_2}=0,064\end{array}\right.\Rightarrow X chứa \left\{\begin{array}{l}N{a}^{+}, K^{+}, B{a}^{2+}\\ Zn{O_2}^{2-}: 0,015\\ O{H}^{-}: 0,034\end{array}\right\}\\ + \left\{\begin{array}{l}N{a}^{+}, K^{+}, B{a}^{2+}\\ Zn{O_2}^{2-}: 0,015\\ O{H}^{-}: 0,034\end{array}\right\}\underset{0,088}{\overset{HCl}{\to }}\left\{\begin{array}{l}N{a}^{+}, K^{+}, B{a}^{2+}\\ C{l}^{-}: 0,088\\ Z{n}^{2+}: \frac{0,088-0,064}{2}=0,012\end{array}\right\}+\underset{0,003 mol}{\underbrace{Zn{\left(OH\right)}_2\downarrow }}\\ \end{array} \\ \Rightarrow m_{Zn{\left(OH\right)}_2}=0,297 gam \end{gathered}$$