When the reaction was carried out in the lab, 944.5 grams potassium chloride reacted with 655.3 grams of oxygen gas. How many grams of the excess reactant are present after the reaction has stopped?

Answer 1

Actually, potassium chloride does not react with oxygen gas, so we can't really answer this question. The potassium is already as oxidized as it can be, and the chlorine in potassium chloride cannot reduce oxygen. Now, if we just want to assume a reaction of 2KCl + 3O₂ ==> 2KClO₃ then we can find the limiting reactant as follows:

moles KCl = 944.5 g x 1 mol KCl/74.6 g = 12.7 moles KCl

moles O₂ = 655.3 g x 1 mol O2/32 g = 20.5 moles O₂

Based on these values and the mol ratio of 2 KCl : 3O₂, the KCl is limiting and O₂ is in excess.

Moles of O₂ used up during the reaction = 12.7 mol KCl x 3 mol O₂/2 mol KCl = 19.1 mol O₂ used up

Moles O₂ left at end of reaction = 20.5 mol - 19.1 mol = 1.4 moles O₂ left over

Mass O₂ left at end of reaction = 1.4 moles O₂ x 32 g/mol = 44.8 g O₂ left