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Stoichiometry Limiting Problem

(Text Book vs. ICE BOX METHOD)

 

Here are the Rules from the other Page

STEP1= SET UP the ICE Box

STEP 2- Find the moles, This is where you have to problem solve.

STEP 3- Find X, find the moles of everything

STEP 4- Answer the questions, convert moles to mass

The question-

You have 20.0 g of elemental sulfur, S, and 160.0 grams of O2. What mass of SO2 can be formed? How much reactant is left over?

S(s) + O2(g) ==> SO2(g)

Tedious Text Book Method- Uses both amounts of reactant and solves for the product. Throws out the larger amount and then reapplies limiting reagent to find the excess. Kent's Simplified Method- Solves for everything in one problem.

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).

  S(s) +  O2(g) ==> SO2(g)
I-Initial      
C-Change -X -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.

 

Moles S

 

20.0g S X

1 mole

=0.623 moles S

32.1g

 

Moles O2

160.0g O2 X

1 mole

=5.000 moles O2

32.0g

 

 

  S(s) +  O2(g) ==> SO2(g) <--there are no moles of products initially
I-Initial

0.623

5.000

  ---------------

C-Change -X -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 S runs out ==>   0.623 mol -X =O ; X is therefore 0.623 mol

if O2 runs out  ==>  5.000 mol -X=O ; X is therefore 5.000 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.)

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

  S(s) +  O2(g) ==> SO2(g)
I-Initial

0.623

5.000

  ---------------

C-Change -0.623 -0.623 +0.623
E-End 0 4.377 mol 0.623 mol

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

What mass of SO2 can be formed?   

0.623mol SO2X

64.1g

=39.9g SO2

1 mol

How much reactant is left over?

4.377mol O2X

32.0g

=140.1g SO2

1 mol

Level 2- The equation now has coefficients.

If you are provided 200.g of sodium and 250. grams of iron(III) oxide, which substance is the limiting reactant? How many grams of Iron are produced? How much reactant is in excess?

6Na    +    Fe2O3 -->   3Na2O +     2Fe

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).

  6Na   +  Fe2O3 ==> 3Na2O +  2Fe
I-Initial        
C-Change -6X -X +3X +2X
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.

 

Moles Na

 

200.g Na X

1 mole

=8.70 moles Na

23.0g

Moles Fe2O3

250.g Fe2O3 X

1 mole

=1.57 moles Fe2O3

159.7g

 

  6Na   +  Fe2O3 ==> 3Na2O +  2Fe
I-Initial

8.70 mol

1.57 mol

--------------

--------------
C-Change -6X -X +3X +2X
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 Na runs out ==>   8.70 mol -6X =O ; X is therefore 1.45 mol

if Fe2O3 runs out  ==>  1.57mol -X=O ; X is therefore 1.57 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.

Na is therefore limiting.

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

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

  6Na   +  Fe2O3 ==> 3Na2O +  2Fe
I-Initial

8.70 mol

1.57 mol

--------------

--------------
C-Change -6(1.45) -(1.45) +3(1.45) +2(1.45)
E-End 0 0.12mol 4.35mol 2.90mol

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

How many grams of Iron are produced?

2.90 mol Fe X

55.85g

=162g Fe

1 mol

How much reactant is in excess?

0.12 mol Fe2O3 X

159.7g

=19.2g SO2

1 mol

 

On to The Stoichiometry of Product Formation and Percent Yield

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