4KCl + 4HNO3 + O2 | ![]() | 4KNO3 + 2Cl2 + 2 H20 |
Michael Feldman
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:
50 kg KCl | formula wt. KCl![]() |
mol KCl | 4 mol KCl = 4 mol KNO3![]() |
mol KNO3 | formula wt. KNO3![]() |
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 | ) |
Copyright © 1997-2010 by Fred Senese
Comments & questions to fsenese@frostburg.edu
Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101/moles/faq/print-limiting-reagent-kcl-kno3.shtml