Đáp án:

 9,9 g

Giải thích các bước giải:

\(\begin{array}{l}
b)\\
ZnC{l_2} + 2NaOH \to Zn{(OH)_2} + 2NaCl\\
{n_{ZnC{l_2}}} = \dfrac{{27,2}}{{136}} = 0,2\,mol\\
{n_{NaOH}} = \dfrac{{600 \times 4\% }}{{40}} = 0,6\,mol\\
{n_{ZnC{l_2}}} < \dfrac{{{n_{NaOH}}}}{2} \Rightarrow NaOH \text{ dư }\\
{n_{Zn{{(OH)}_2}}} = {n_{ZnC{l_2}}} = 0,2\,mol\\
{n_{NaOH}}  \text{ dư }= 0,6 - 0,2 \times 2 = 0,2\,mol\\
Zn{(OH)_2} + 2NaOH \to N{a_2}Zn{O_2} + 2{H_2}O\\
{n_{Zn{{(OH)}_2}}} > \dfrac{{{n_{NaOH}}} \text{ dư }}{2} \Rightarrow Zn{(OH)_2} \text{ dư }\\
{n_{Zn{{(OH)}_2}}} \text{ dư }= 0,2 - \dfrac{{0,2}}{2} = 0,1\,mol\\
{m_{Zn{{(OH)}_2}}}  \text{ dư }= 0,1 \times 99 = 9,9g
\end{array}\)