1. Unit 10: Stoichiometry Limiting reactant calculations

2. 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 calculations will be used. The L.R. is the one that yields the smaller amount.

3. A real-world example: Making Funfetti Cupcakes! ___mix + ___eggs + ___oil  ___pan of cupcakes Let’s say you have… 1321 3 mixes 15 eggs 8 tbsp oil x x x 1 mix 1 pan = = 3 eggs 1 pan 2 tbsp oil 1 pan = 3 pans 5 pans 4 pans Mixes are the “limiting reactant” because they are used up first! Ok… Time to relate this back to chemistry…

4. Example: If 17.1g of potassium reacts with 14.3g 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: 2K + F 2  2KF 17.1gK 1 x 2 mol KF 2 mol K = 25.4g KF 1 mol K 39.10g K xx 58.10gKF 1 mol KF 14.3gF 2 ?gKF 1K=39.10 1F=19.00 58.10g 14.3gF 2 1 2 mol KF 1 mol F 2 = 43.7g KF 1 mol F 2 38.00gF 2 x 58.10gKF 1 mol KF x xx Potassium is the L.R. 25.4g of potassium fluoride can be produced. K: 17.1gK U: ?gKF K: 14.3gF 2 U: ?gKF

5. Unit 10: Stoichiometry Excess reactant calculations

6. 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? Word equation: Formula Equation: iron + oxygen  iron (III) oxide Fe+O2O2  Fe 2 O 3 32 4 Fe +3 O -2 13.17 mol Fe 1 = 6.585 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe x 18.19 mol O 2 1 = 12.13 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2 x Can be produced: K: 13.17 molFe U: ? molFe 2 O 3 K: 18.19 molO 2 U: ? molFe 2 O 3

7. Example: Limiting Reactant AND Excess Reactant Calculations a)What mass of CoCl 3 is formed from the reaction of 3.478x10 23 atoms Co with 57.92L of Cl 2 gas at STP? b)How much of the excess reactant reacts and how much is left over? 2Co + 3Cl 2  2CoCl 3 3.478x10 23 atomsCo 1 x 2 mol CoCl 3 2 mol Co = 95.40gCoCl 3 1 mol Co 6.02x10 23 atomsCo x 165.13gCoCl 3 1 mol CoCl 3 1Co=58.78 3Cl=106.35 165.13g 284.7gCoCl 3 x 57.92L Cl 2 1 x 2 mol CoCl 3 3 mol Cl 2 = 1 mol Cl 2 22.4L Cl 2 x 165.13gCoCl 3 1 mol CoCl 3 x LR ER Can be produced: K: 3.478x10 23 atomsCo U: ? gCoCl 3 K: 57.92 LCl 2 U: ? gCoCl 3

8. Example: Limiting Reactant AND Excess Reactant Calculations a)What mass of CoCl 3 is formed from the reaction of 3.478x10 23 atoms Co with 57.92L of Cl 2 gas at STP? b)How much of the excess reactant reacts and how much is left over? 2Co + 3Cl 2  2CoCl 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. 19.41L Cl 2 3.478x10 23 atomsCo 1 x 3 mol Cl 2 2 mol Co = 1 mol Co 6.02x10 23 atomsCo x 22.4L Cl 2 1 mol Cl 2 x LRER K: 3.478x10 23 atomsCo U: ? L Cl 2 Reacts:

9. Example: Limiting Reactant AND Excess Reactant Calculations a)What mass of CoCl 3 is formed from the reaction of 3.478x10 23 atoms Co with 57.92L of Cl 2 gas at STP? b)How much of the excess reactant reacts and how much is left over? 2Co + 3Cl 2  2CoCl 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). 38.51L Cl 2 LRER 57.92L Cl 2 -19.41L Cl 2 Left over: