1. Limiting Reagents/Reactants

2. Limiting Reactant- the reactant that is completely consumed in the chemical reaction Excess Reactant- the reactant that is present in more than required quantities in the chemical reaction When the limiting reactant is used up, the reaction will stop. No more product can be made, regardless of how much of the excess reactant may be present. Therefore, the limiting reactant determines how much product is produced

3. Real Life Examples Washing dishes – you can have too much, or too little soap or grease Sugar in blood levels in your body. Too much sugar you are tired, too little, have no energy Salt & Potassium in body. Muscle cramps & thirst

4. Solving Limiting Factor Problems When you are given amounts of two or more reactants to solve a stoichiometric problem, you must identify the limiting reactant

5. How to identify the limiting reactant: Problem type: What mass of product C could be obtained when mass A is reacted with mass B? Chemical reaction format: 2A+ B  3C + D

6. Problem Solving Strategy Mol A Mass C Mol C Mass A Mass B Mol A Mol B 2A+ B  3C + D Note: 2 masses given Use mole ratio to determine the limiting factor (e.g. A is limiting) The limiting factor always controls the amount of product being produced Mole ratio

7. Example 1: A reaction contains 134.9g of Aluminium and 96.0g of oxygen. Determine the excess & limiting reagent and the amount of product formed. Al + O 2  Al 2 O 3 4Al + 3O 2  2Al 2 O 3 Step 1: Write out the balanced chemical equation

8. Step 2: Fill in chart with information you know Balanced Equation 4Al3O 2 2Al 2 O 3 Mole Ratio 432 Mass (m) 134.9g96.0g Molar Mass (M) 26.98g/mol 32.0g/mol Moles (n)

9. Step 3: Convert given mass into moles for both (n=m/M) Moles of O 2 = 96.0g 32.0g/mol = 4.99 mol of Al Moles of Al = 134.9g 27.0g/mol = 3.00 mol of O 2

10. Fill in chart with information you know Balanced Equation 4Al3O 2 2Al 2 O 3 Mole Ratio 432 Mass (m) 134.9g96.0g ?g Molar Mass (M) 26.98g/mol 32.0g/mol Moles (n)4.99mol3.00mol

11. Step 4: Take the moles of each reactant and using the mole ratio, determine how much product would be made The O 2 limiting reactant because it produces the least amount of moles of product (Al 2 O 3 ) Mols of Al 2 O 3 produced from O 2 3.00 mol O 2 2 mol Al 2 O 3 3 mol O 2 X = 2.00 mol Al 2 O 3 Mols of Al 2 O 3 produced from Al 4.99 mol Al 2 mol Al 2 O 3 4 mol Al X= 2.50 mol Al 2 O 3

12. Fill in chart with information you know Balanced Equation 4Al3O 2 2Al 2 O 3 Mole Ratio 432 Mass (m) 134.9g96.0g ?g Molar Mass (M) 26.98g/mol 32.0g/mol102g/mol Moles (n)4.99 mol3.00mol2.0mol

13. Step 5: Convert moles of required substance to the mass (m=n x M) mass of Al 2 O 3 = 2.00 mol of Al 2 O 3 X 102g/mol = 204g of Al 2 O 3

14. Example 2: Pentane (C 5 H 12 ) is a major component of gasoline. What mass of water would be produced when 28.5 g of pentane reacts with 3.00 g of oxygen gas? 1C 5 H 12 + 8O 2  6H 2 O + 5CO 2 Reactants Products m= 28.5gm= 3.00gm= ?g

15. Step 2: Fill in chart with information you know Balanced Equation 1C 5 H 12 8O 2 6H2O6H2O5CO 2 Mole Ratio 186 5 Mass (m) 28.5g3.00g Molar Mass (M) 72.17g/mol 32g/mol Moles (n)

16. Step 3: Convert given mass into moles for both (n=m/M) Moles of O 2 = 3.00g 32g/mol = 0.3958 mols of C 5 H 12 Moles of C 5 H 12 = = 0.09375 mol of O 2 28.5g 72g/mol

17. Fill in chart with information you know Balanced Equation 1C 5 H 12 8O 2 6H2O6H2O5CO 2 Mole Ratio 186 5 Mass (m) 28.5g3.00g Molar Mass (M) 72.17g/mol 32g/mol Moles (n) 0.39580.09375

18. Step 4: Take the moles of each reactant and using the mole ratio, determine how much product would be made The O 2 limiting reactant because it produces the least amount of moles of product (H 2 O) Mols of H 2 O produced from O 2 0.09375 mol O 2 6 mol H 2 O 8 mol O 2 X = 0.703 mol H 2 O Mols of H 2 O produced from C 5 H 12 0.3958 mol C 5 H 12 6 mol H 2 O 1 mol C 5 H 12 X = 2.37 mol H 2 O

19. Fill in chart with information you know Balanced Equation 1C 5 H 12 8O 2 6H2O6H2O5CO 2 Mole Ratio 186 5 Mass (m) 28.5g3.00g Molar Mass (M) 72.17g/mol 32g/mol 18.02g/mol Moles (n) 0.093750.3958 0.703 mols

20. Step 5: Convert moles of required substance to the mass (m=n x M) mass of H 2 O = 0.0703 mols H 2 O X 18.02g/mol = 1.27g of H 2 O

21. Example 3: Calculate the mass of aluminium chloride that can be produced from 20.0 g of aluminium and 30.0 g of chlorine gas. Step 1: Write the balanced equation for the reaction. 2 Al (s) + 3 Cl 2 (g)  2 AlCl 3 (s)

22. Step 2: Calculate the number of moles of each reactant. (Using n=m/M) 2 Al (s) + 3 Cl 2 (g)  2 AlCl 3 (s) m 20.0 g 30.0g M 27.0g/mol 2(35.5) =71.0g/mol n = 0.741 mol Al =0.423 mol Cl 2 20.0g 27.0g/mol 30.0g 71.0 g/mol

23. Step 3: Take the moles of each reactant and using the mole ratio, determine how much product would be made The Cl 2 limiting reactant because it produces the least amount of moles of product (AlCl 3 ) Mols of AlCl 3 produced from Al 0.741 mol Al 2 mol AlCl 3 2 mol Al X = 0.741 mol AlCl 3 Mols of AlCl 3 produced from Cl 2 0.423 mol Cl 2 2 mol AlCl 3 3 mol Cl 2 X = 0.282 mol AlCl 3

24. Step 4. Calculate the number of moles using the limited reagent. Moles of AlCl 3 = nM = 0.423mol(133.33g/mol) = 56.4 g of AlCl 3