Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl

(ii) From O₂

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used

(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
They have different number of Neutrons and protons, so their masses are different
Hope this helps!
Because this classification contrasts with that of crystalline solids whose atoms are arranged in a regular and orderly fashion forming crystalline networks.