For the following reaction, 28.6 grams of zinc oxide are allowed to react with 9.54 grams of water . zinc oxide(s) water(l) zinc hydroxide(aq) What is the maximum mass of zinc hydroxide that can be formed

34.9 g of Zn(OH)₂ is the maximum mass that can be formed

Explanation:

Let's state the reaction:

ZnO(s)  + H₂O(l) → Zn(OH)₂ (aq)

First of all, we need to determine the moles of each reactant and state the limiting:

28.6 g . 1mol /81.38 g = 0.351 moles of ZnO

9.54 g . 1mol /18 g = 0.53 moles of water

As ratio is 1:1, for 0.53 moles of water, we need 0.53 moles of ZnO, but we only have 0.351, so the limiting reactant is the ZnO.

Ratio with the product is also 1:1. From 0.351 moles of oxide we can produce 0.351 moles of hydroxide. Let's calculate the mass:

0.351 mol . 99.4 g /1mol = 34.9 g

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