Mostly the electrons will determine the reactivity
<span>
</span>

<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3
So, for 2.3 ✕ 10−6 m OH-
[Cu2+] =

%yield = 54.6%
<h3>Further explanation</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
(theoretical)
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
<h3 />
Reaction
2ZnS+3O₂ ⇒ 2ZnO+2SO₂
MW ZnS = 97.474 g/mol

MW ZnO = 81.38 g/mol
- mol ZnO (from mol ZnS as limiting reactant, O₂ excess)


Theoretical production = 125.388
