Lời giải:\(\begin{align} & {{n}_{ZnS{{O}_{4}}}}=x\,\,mol \\ & {{n}_{Zn{{(OH)}_{2}}\,\,(1)}}=3y\,\,\,mol \\ & {{n}_{Zn{{(OH)}_{2\,\,(2)}}}}=2y\,\,\,mol \\ & {{n}_{K\text{O}{{H}_{(1)}}}}=0,22\,\,\,mol \\ & {{n}_{K\text{O}{{H}_{(2)}}}}=0,28\,\,\,mol \\ & Z{{n}^{2+}}\,\,\,\,\,\,+\,\,\,\,\,\,2\text{O}{{H}^{-}}\,\,\,\,\to \,\,\,\,\,Zn{{(OH)}_{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \\ & Zn{{(OH)}_{2}}\,\,\,\,+\,\,\,\,2\text{O}{{H}^{-}}\,\,\,\to \,\,\,\,ZnO_{2}^{2-}\,\,\,\,+\,\,\,\,2{{H}_{2}}O\,\,\,\,\,\,\,\,\,\,\,\,\,(2) \\ \end{align}\)
Để làm tan y mol kết tủa thì phải cần số mol OH- là: 0,28 – 0,22 = 0,06
Vậy: y = 0,03
Trong TN2 thì phản ứng (2) có xảy ra nên:
\(\begin{align} & {{n}_{OH_{(2)}^{-}}}=4.{{n}_{Z{{n}^{2+}}}}-2.{{n}_{Zn{{(OH)}_{2}}}} \\ & \Leftrightarrow 0,28=4\text{x}-2.2.0,03 \\ & \Leftrightarrow x=0,1\,\,\,mol \\ & m=16,10\,\,\,gam \\ \end{align}\)
