Limiting reagents
Definition:
The limiting reagent is the reactant that is completely used up in a chemical reaction and therefore determines and limits the amount of product being formed.
There are two methods to determine the limiting reagent.
Method 1: comparing mole ratios
- Balance the chemical equation.
- Convert all given information into moles.
- Calculate the mole ratio from the given information. Compare the calculated ratio to the mole ratio from the balanced chemical equation to determine the limiting and excess reagent.
- Use the amount of limiting reactant to calculate the amount of product produced.
- If necessary, subtract the mass of excess reagent consumed from the total mass of excess reagent given to find the amount of remaining excess reactant.
Example:
C6H12O6 + O2 → CO2 + H2O + energy
Calculate the amount of carbon dioxide produced in the reaction of 25.0 grams of glucose with 40.0 grams of oxygen?
1. Balance the chemical equation.
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + energy
2. Convert all given information into moles
25.0g C6H12O6 x 1 mol = 0.1389 mol C6H12O6
180g
40.0g O2 x 1 mol = 1.250 mol O2
32g
3. Calculate the mole ratio from the given information. Compare the calculated ratio to the mole ratio from the balanced chemical equation to determine the limiting and excess reagent
Calculated mole ratio:
1.250 mol O2 = 8.999 mol O2
0.1389 mol C6H12O6 1 mol C6H12O6
Mole ration from balanced chemical equation:
6 mol O2
1 mol C6H12O6
Conclusion: C6H12O6 is the limiting reagent while O2 is the excess reagent
4. Use the amount of limiting reactant to calculate the amount of product produced. Round to proper number of significant figures.
0.1389 mol C6H12O6 x 6 mol CO2 = 0.834 mol CO2
1 mol C6H12O6
5. Subtract the mass of excess reagent consumed from the total mass of excess reagent given to find the amount of remaining excess reactant.
1.250 mol O2 - 0.834 mol O2 = 0.417 moles of oxygen left over
Method 2: calculating and comparing the amount of product each reactant will produce
- Balance the chemical equation.
- Using stoichiometry, calculate the amount of product produced by each individual reactant.
- Determine which is limiting and excess reagent. The limiting reagent determines the amount of product that can be produced.
- If necessary, subtract the mass of excess reagent consumed from the total mass of excess reagent given to find the amount of remaining excess reactant.
Example:
Mg +O2 → MgO
Calculate the mass of magnesium oxide possible if 2.40g Mg reacts with 10.0g O2
1. Balance the chemical equation.
2 Mg + O2 → 2 MgO
2. Using stoichiometry, calculate the amount of product produced by each individual reactant.
Amount of MgO produced from 2.40g of Mg:
2.40g Mg x 1 mol Mg x 2 mol MgO x 40.3g MgO = 3.98g MgO
24.3g Mg 2 mol Mg 1 mol MgO
Amount of MgO produced from 10.0g O2:
10.0g O2 x 1 mol O2 x 2 mol MgO x 40.3g MgO = 25.2g MgO
32g O2 1 mol O2 1 mol MgO
3. Determine which is limiting and excess reagent. The limiting reagent determines the amount of product that can be produced.
Mg produces less amount of MgO than O2; therefore Mg is the limiting reagent.
O2 produces more amount of MgO than Mg; therefore O2 is the excess reagent.
In conclusion: 3.98g of MgO would be produced if 2.40g Mg reacts with 10.0g O2
4. Subtract the mass of excess reagent consumed from the total mass of excess reagent given to find the amount of remaining excess reactant.
2.40g Mg x 1 mol Mg x 1 mol O2 x 32g O2 = 1.58g O2
24.3g Mg 2 mol Mg 1 mol O2
10.0g – 1.58g = 8.42g oxygen left over.
work sheets with answers
http://misterguch.brinkster.net/pra019.pdf
http://www.chemteam.info/Stoichiometry/WS-limiting-reagent.html
http://www.chemistry.wustl.edu/~coursedev/Online%20tutorials/Plink/limreagkey.htm
http://science.widener.edu/svb/tutorial/limitreagentcsn7.html
http://www.chemteam.info/Stoichiometry/WS-limiting-reagent.html
http://www.chemistry.wustl.edu/~coursedev/Online%20tutorials/Plink/limreagkey.htm
http://science.widener.edu/svb/tutorial/limitreagentcsn7.html
more resources
Videos:
http://chemcollective.org/activities/tutorials/stoich/limiting-reagents
http://www.youtube.com/watch?v=IvCPLCQ-YK0
http://www.youtube.com/watch?v=rESzyhPOJ7I
Analogies:
http://resources.educ.queensu.ca/science/main/concept/chem/c02/C02TPGC2.htm
Visual representation:
http://www.chem.queensu.ca/courses/yr1resources/subsite/LimitingReagent.html
Other sites explaining limiting reagents:
http://www.ausetute.com.au/exceslim.html
http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.html
http://scidiv.bellevuecollege.edu/bg/limiting.html
http://chemcollective.org/activities/tutorials/stoich/limiting-reagents
http://www.youtube.com/watch?v=IvCPLCQ-YK0
http://www.youtube.com/watch?v=rESzyhPOJ7I
Analogies:
http://resources.educ.queensu.ca/science/main/concept/chem/c02/C02TPGC2.htm
Visual representation:
http://www.chem.queensu.ca/courses/yr1resources/subsite/LimitingReagent.html
Other sites explaining limiting reagents:
http://www.ausetute.com.au/exceslim.html
http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.html
http://scidiv.bellevuecollege.edu/bg/limiting.html