Answer:
6.347 g Zinc to the nearest thousandth.
Explanation:
Using the molar atomic mass of the elements:
65.38 g Zinc produces 65.38 + 2(16 + 1.008) = 99.396 g Zn(OH)2
So by proportion amount of zinc required to produce 9,65 g Zn(OH)2
= (65.38/99.396) * 9.65 g Zinc
= 0.65777 * 9.65
= 6.347g Zinc.