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
Answer:
What give me a few minutes I have to finish my test them I will answer in comments.
Oxygen because it is on the left of the periodic table so it has a strong pull.
Answer: four hundred and 18
Explanation:
Simple math showed that only 16 words are possible from a two-letter combination, but a three-letter code produces 64 words. Operating on the principle that the simplest solution is often correct, researchers assumed a three-letter code called a codon.