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Limiting Stoichiometry Problems

(Ice Box Method)

 

This technique is used in later units of this course, but when modified, it really works nicely here.

What is a limiting Reactant?

The Limiting reactant runs out first and limits the amount of product that can be made.

What is the excess reactant?

The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed.  The excess reactant remains because there is nothing with which it can react.

 Car example

[Image]

No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made.  Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up.

What about the spares....?

There a few ways to do this. Many are overly complicated. This is very straight forward and can solve for every compound all in one shot.

Text book video link==> Limiting Reagent Concept Tutorial

Typical starter question-

 How many moles of H2 is produced if there is 0.30 mol of Zn and 0.52 moles of HCl?  How many moles of reactant are left over?

Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g)

 

STEP1= SET UP the ICE Box  below the balanced reaction.   (I-Initial  C-Change E-End). Using the coefficients, (the #'s in front of each compound) to determine the relative change. If there is a 2 in front, use 2X, if there is a 3 use 3X. Also reactants decrease in their amounts (-X), products will increase (+X).

  Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g)
I-Initial        
C-Change -X -2X +X +X
E-End        

Step 2- Moles are the only thing allowed in the ice box. Find the moles of each compound and insert that value in its I box.

How nice I don't have to convert to moles first. They gave it to me. 0.30 mol of Zn and 0.52 moles of HCl.

  Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g) <--there are no moles of products initially
I-Initial 0.30 mol 0.52 mol ----- -----
C-Change -X -2X +X +X
E-End        

Step 3- Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction.

if Zn runs out ==>   0.30 mol -X =O ; X is therefore 0.30 mol

if HCl runs out  ==>  0.52 mol -2x=O ; X is therefore 0.26 mol

Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX.

(If you applied the incorrect, larger X value, will get a negative amount at the end, so go back and change it.)

  Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g)
I-Initial 0.30 mol 0.52 mol ------- -------
C-Change -0.26 -2(0.26) +0.26 +0.26
E-End        

Now solve for everything (add or subtract down each column).

  Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g)
I-Initial 0.30 mol 0.52 mol ------- -------
C-Change -0.26 -2(0.26) +0.26 +0.26
E-End 0.04 mol  0 0.26 mol 0.26 mol

Step 4- Answer the questions, you have them moles of everything.

(You may have to convert these to grams for other types of limiting problems.)

How many moles of H2 is produced if there is 0.30 mol of Zn and 0.52 moles of HCl?  0.26 mol

How many moles of reactant are left over? 0.04 mol

Now on to Limiting Stoichiometry using Masses

Chemical Demonstration Videos