Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

what is the limiting reactant? (TYPE EITHER KCl or O2)

what is the excess reactant? (TYPE EITHER KCl or O2)

how many moles of the excess reactant will be left over? moles (TYPE just the number to the correct amount of significant figures)

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

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error