DO NOW What mass of aluminum hydroxide is

DO NOW: What mass of aluminum hydroxide is needed to decompose in order to produce aluminum oxide and 35 L of water at STP? Word equation: Formula equation:

After today you will be able to… • Calculate the limiting reactant using two mass-mass calculations

Limiting Reactant Calculations The limiting reactant (L. R. ) is the reactant which runs out first and limits the amount of product that can be made. • Two mass-mass calculations will be used. • The L. R. is the one that yields the smaller amount.

Limiting Reactant Example

A real-world example: Making Funfetti Cupcakes! 1 Ok… Time Mixes are the “limiting to relate reactant” because they are used up first! this back to chemistry… ___mix 3+ ___eggs + 2___oil ___pan of cupcakes 1 Let’s say you have… 3 mixes x 15 eggs x 8 tbsp oil x 1 pan 1 mix 1 pan 3 eggs 1 pan 2 tbsp oil = 3 pans = 5 pans = 4 pans

Example: If 17. 1 g of potassium reacts with 14. 3 g of fluorine, which reactant is the limiting reactant and what mass of potassium fluoride can theoretically be produced? Word equation: potassium + fluorine potassium fluoride Formula Equation: 2 K K: 17. 1 g. K U: ? g. KF 17. 1 g. K + F 2 14. 3 g. F 2 17. 1 g K x 1 mol K x 2 mol KF x 58. 10 g. KF = 39. 10 g K 1 2 mol K 1 mol KF K: 14. 3 g. F 2 U: ? g. KF 14. 3 g. F 2 x 1 mol F 2 x 2 mol KF x 58. 10 g. KF = 38. 00 g. F 2 1 mol KF 1 2 KF ? g. KF 1 K=39. 10 1 F=19. 00 58. 10 g 25. 4 g KF 43. 7 g KF Potassium is the L. R. 25. 4 g of potassium fluoride can be produced.

After today you will be able to… • Calculate the limiting reactant using other mole relationships • Calculate how much of the excess reactant reacts and how much is left over

Other L. R. Calculations Example: How many moles of iron (III) oxide can be produced from the reaction of 13. 17 moles of iron with 18. 19 moles of oxygen? Fe+3 O-2 Word equation: iron + oxygen iron (III) oxide Formula Equation: 4 Fe + 3 O 2 2 Fe 2 O 3 K: 13. 17 mol. Fe U: ? mol. Fe 2 O 3 Can be produced: 13. 17 mol Fe x 2 mol Fe 2 O 3 = 6. 585 mol Fe 2 O 3 4 mol Fe 1 K: 18. 19 mol. O U: ? mol. Fe 2 O 3 2 18. 19 mol O 2 x 2 mol Fe 2 O 3 = 12. 13 mol Fe 2 O 3 3 mol O 1 2

Example: Limiting Reactant AND Excess Reactant Calculations a) What mass of Co. Cl 3 is formed from the reaction of 3. 478 x 1023 atoms Co with 57. 92 L of Cl 2 gas at STP? b) How much of the excess reactant reacts and how much is left over? 1 Co=58. 78 K: 3. 478 x 1023 atoms. Co U: ? g. Co. Cl 3 LR 2 Co + 3 Cl 2 2 Co. Cl 3 3 Cl=106. 35 165. 13 g 1 mol Co x 2 mol Co. Cl 3 x 165. 13 g. Co. Cl 3 3. 478 x 1023 atoms. Co x = 1 mol Co. Cl 3 6. 02 x 1023 atoms. Co 2 mol Co 1 K: 57. 92 LCl 2 U: ? g. Co. Cl 3 ER Can be produced: 57. 92 L Cl 2 x 1 mol Cl 2 x 2 mol Co. Cl 3 x 165. 13 g. Co. Cl 3 = 22. 4 L Cl 2 3 mol Cl 2 1 1 mol Co. Cl 3 95. 40 g. Co. Cl 3 284. 7 g. Co. Cl 3

Example: Limiting Reactant AND Excess Reactant Calculations a) What mass of Co. Cl 3 is formed from the reaction of 3. 478 x 1023 atoms Co with 57. 92 L of Cl 2 gas at STP? b) How much of the excess reactant reacts and how much is left over? LR ER 2 Co + 3 Cl 2 2 Co. Cl 3 To find out how much excess reactant reacts, do a third calculation using the limiting reactant as the known, and the excess reactant as the unknown. K: 3. 478 x 1023 atoms. Co U: ? L Cl 2 3. 478 x 1023 atoms. Co x 1 mol Co x 3 mol Cl 2 x 22. 4 L Cl 2 = 1 6. 02 x 1023 atoms. Co 2 mol Co 1 mol Co. Cl 3 Reacts: 19. 41 L Cl 2

Example: Limiting Reactant AND Excess Reactant Calculations a) What mass of Co. Cl 3 is formed from the reaction of 3. 478 x 1023 atoms Co with 57. 92 L of Cl 2 gas at STP? b) How much of the excess reactant reacts and how much is left over? LR ER 2 Co + 3 Cl 2 2 Co. Cl 3 To find out how much is left over (unreacted), subtract the amount of the excess reactant that reacted from the original amount of excess reactant (from the problem). 57. 92 L Cl 2 -19. 41 L Cl 2 Left over: 38. 51 L Cl 2

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