Home

Home
Common Compounds
Exam Guide
FAQ
Features
Glossary
Construction Kits
Companion Notes
Just Ask Antoine!
Simulations
Slide Index
Toolbox
Tutorial Index

FAQ
Introduction
Measurement
Matter
Atoms & ions
Compounds
Chemical change
The mole
Gases
Energy & change
The quantum theory
Electrons in atoms
The periodic table
Chemical bonds
Solids
Liquids
Solutions
Acids & bases
Redox reactions
Reaction rates
Organic chemistry
Everyday chemistry
Inorganic chemistry
Environmental chemistry
Laboratory
History of chemistry
Miscellaneous


Home :FAQ :The mole conceptPrint | Comment
Previous Question Next Question

How can amount of product (KNO3) be predicted from amounts for two reactants (KCl, HNO3)?

How many kilograms of potassium nitrate will be produced from 50.0 kg of potassium chloride and 50.0 kg of nitric acid in the following reaction?

4KCl + 4HNO3 + O2rightarrow4KNO3 + 2Cl2 + 2 H20


Michael Feldman

Whenever you have the amount of both reactants, you have a limiting reactant problem. For example, if I have 10 slices of bread and 10 slices of swiss cheese, I can make 5 sandwiches at most (I insist that a sandwich must have 2 slices of bread and at least one slice of cheese.) The amount of bread I have limits the amount of product I get.

You'll need 4 moles of KCl for every 4 moles of nitric acid, or some of your starting materials will be wasted. The reactant that is present in the smallest molar amount will limit the amount of product you get. So your overall strategy is to find which reactant (the KCl or nitric acid) gives you the lowest yield of potassium nitrate. That's the amount of potassium nitrate you'll actually get.

In a little more detail, the plan is to:

  1. Find how many kilograms of KNO3 you get from 50 kg of KCl. You need to convert an amount of one substance (KCl) into an amount of another (KNO3, so you need a mole-to-mole conversion factor to bridge the two amounts. From the equation, you have 4 mol KCl = 4 mol KNO3. The strategy is
    50 kg KCl formula wt. KCl
    longrightarrow
    mol KCl 4 mol KCl = 4 mol KNO3
    longrightarrow
    mol KNO3 formula wt. KNO3
    longrightarrow
    kg KNO3

    and the actual setup for this part of the problem is
    50 kg KCl (1000 g KCl
    1 kg KCl
    ) (1 mol KCl
    74.55 g KCl
    ) (4 mol KNO3
    4 mol KCl
    ) (101.11 g KNO3
    1 mol KNO3
    ) (1 kg KNO3
    1000 g KNO3
    )

  2. Find how many kilograms of KNO3 you get from 50 kg of nitric acid. This calculation is very similar to the one in the previous step.
  3. Report the lower of the two predictions as your theoretical yield of KNO3. The reactant that limits the amount of product (the "limiting reactant") will be completely consumed in the reaction. There will be some of the other reactant left over after the reaction (which will contaminate the product).

Author: Fred Senese senese@antoine.frostburg.edu



General Chemistry Online! How can amount of product (KNO_3_) be predicted from amounts for two reactants (KCl, HNO_3_)?

Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/moles/faq/limiting-reagent-kcl-kno3.shtml