Limiting Reagents

Limiting reagents are another type of stoichiometry problem. More complicated compared to the earlier stoichiometry problems, limiting reagent problems require the determination of the limiting reagent. These types of problems can be easily recognized by the information provided in the problem.

Lets look at how the limiting reagent problems are generally worded compared to typical stoichiometry problem.

Limiting Reagent Problem;

In the reaction

2Hg(l) + O2(g) ---> 2HgO(s)

20.0 g of Hg are reacted with 5.00 g of O2. Calculate the maximum amount, in grams, of HgO formed.

Typical stoichiometry Problem;

In the reaction

2Hg(l) + O2(g) ---> 2HgO(s)

Calculate the amount, in grams, of HgO formed when 20.0 g of Hg react with excess O2.

In the typical stoichiometry problem the reagent in excess is given, not so for limiting reagent problems.

 

So how do we solve limiting reagent problems? Let's do one! Click here if you would like to see this problem worked out from class.

In the reaction

2C2H6(g) + 7O2(g) ---> 4CO2(g) + 6H2O(g)

Determine the mol of carbon dioxide formed when 2.00 mol of ethane are reacted with 10.0 mol of oxygen.

To solve this limiting reagent problem I recommend using a table approach which I have developed over the years. To solve this problem we prepare a table which looks like the following;

(mol reactant A)o

(mol reactant B)required

(mol reactant B)o

Conclusion

 

 

 

 

To fill out the table we arbitrarily select one of the reactant (it does not make a difference which one) as reactant A. Once reactant A is selected reactant B is defined. So if we select reactant A as C2H6, the reactant B must be O2. So the table would look like;

(mol reactant C2H6)o

(mol reactant O2)required

(mol reactant O2)o

Conclusion

 

 

 

 

Now we can add the mol of each reactant initially and the mol of O2 required. according to the statement of the problem we are given 2.00 mol of C2H6 and 10.0 mol of O2 initially.

(mol reactant C2H6)o

(mol reactant O2)required

(mol reactant O2)o

Conclusion

2.00

 

10.0

 

To determine the mol of O2 required, we must answer the question how many mol of O2 are required to react with 2.00 mol of C2H6. So the following calculation;

This calcualtion says 7.00 mol of O2 are required to completely react with 2.00 mol of C2H6.

So the table would now look like,

(mol reactant C2H6)o

(mol reactant O2)required

(mol reactant O2)o

Conclusion

2.00

7.00

10.0

 

To determine the limiting reagent we compare the mol of O2 required to mol of O2 initially. In this case 7.00 mol O2 are required and we have 10.0 mol O2 initially. So we have initially more mols of O2 than are required. So O2 is in excess and C2H6 is limiting.

(mol reactant C2H6)o

(mol reactant O2)required

(mol reactant O2)o

Conclusion

2.00

7.00

10.0

O2 is excess

C2H6 is limiting

Now we use the limiting reagent to calculate the amount of CO2 formed.

Here is a second limiting reagent problem. Click here if you would like to see this problem worked out from class. Note the dropdown menu Chapter titles in the movie timeline that allow you to jump to certain key discussion points during the video.

Calculate the mass calcium nitride formed when 50.0 g of calcim are combined with 50.0 g of nitrogen according to the equation;

3Ca(s) + N2(g) ---> Ca3N2(s)

To solve this problem we must build our table. We will start with Ca in our table;

(mol reactant Ca)o

(mol reactant N2)required

(mol reactant N2)o

Conclusion

 

 

 

 

We need to calculate the mols of each reactant;

So the table becomes;

(mol reactant Ca)o

(mol reactant N2)required

(mol reactant N2)o

Conclusion

1.25

 

1.79

 

Now we must calculate how many mols of N2 are required to react with 1.25 mols Ca;

So the table becomes;

(mol reactant Ca)o

(mol reactant N2)required

(mol reactant N2)o

Conclusion

1.25

0.412

1.79

 

Comparing the mols of N2 required to the mols of N2 initially we see there is an excess of N2 available. Only 0.412 mols N2 are required and we were given 1.79 mols N2. Therefore,

(mol reactant Ca)o

(mol reactant N2)required

(mol reactant N2)o

Conclusion

1.25

0.412

1.79

N2 is excess

Ca is limiting

So we will use the mols of Ca (the liming reagent) to calculate the mass of calcium nitride formed;

 

So would you like to try a problem on your own?

Calculate the mass of SbI3 formed when 1.20 g of Sb react with 2.40 g of I2, according to the equation,

2Sb(s) + 3I2(s) ---> 2SbI3(s)

Answer

If you are not sure of the answer let's step through the problem.