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Unit 10: Stoichiometry

Unit 10: Stoichiometry. Limiting reactant 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 calculations will be used. The L.R. is the one that yields the smaller amount.

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Unit 10: Stoichiometry

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

  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 + F2 2KF 1K=39.10 1F=19.00 17.1gK 14.3gF2 ?gKF K: 17.1gK 58.10g U: ?gKF 1 mol K 39.10g K 17.1g K 1 2 mol KF 2 mol K 58.10gKF 1 mol KF x x x x = 25.4g KF K: 14.3gF2 U: ?gKF 1 mol F2 38.00gF2 14.3gF2 1 2 mol KF 1 mol F2 58.10gKF 1 mol KF x x x = 43.7gKF Potassium is the L.R. 25.4g of potassium fluoride can be produced.

  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: Fe+3 O-2 iron + oxygen  iron (III) oxide 4 3 2  Fe2O3 Fe + O2 K: 13.17 molFe U: ? molFe2O3 Can be produced: 2 mol Fe2O3 4 mol Fe 13.17 mol Fe 1 x 6.585 mol Fe2O3 = K: 18.19 molO2 U: ? molFe2O3 2 mol Fe2O3 3 mol O2 18.19 mol O2 1 x 12.13 mol Fe2O3 =

  7. Example: Limiting Reactant AND Excess Reactant Calculations What mass of CoCl3 is formed from the reaction of 3.478x1023 atoms Co with 57.92L of Cl2 gas at STP? How much of the excess reactant reacts and how much is left over? 2Co + 3Cl2  2CoCl3 1Co=58.78 3Cl=106.35 K: 3.478x1023 atomsCo 165.13g U: ? gCoCl3 LR 165.13gCoCl3 1 mol CoCl3 1 mol Co 6.02x1023atomsCo 2 mol CoCl3 2 mol Co 3.478x1023atomsCo 1 x x x = K: 57.92 LCl2 95.40gCoCl3 Can be produced: U: ? gCoCl3 ER 1 mol Cl2 22.4L Cl2 2 mol CoCl3 3 mol Cl2 57.92L Cl2 1 165.13gCoCl3 1 mol CoCl3 x x x 284.7gCoCl3 =

  8. Example: Limiting Reactant AND Excess Reactant Calculations What mass of CoCl3 is formed from the reaction of 3.478x1023 atoms Co with 57.92L of Cl2 gas at STP? How much of the excess reactant reacts and how much is left over? 2Co + 3Cl2  2CoCl3 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. LR ER K: 3.478x1023atomsCo U: ? L Cl2 3 mol Cl2 2 mol Co 22.4L Cl2 1 mol Cl2 3.478x1023atomsCo 1 1 mol Co 6.02x1023atomsCo x x x = 19.41LCl2 Reacts:

  9. Example: Limiting Reactant AND Excess Reactant Calculations What mass of CoCl3 is formed from the reaction of 3.478x1023 atoms Co with 57.92L of Cl2 gas at STP? How much of the excess reactant reacts and how much is left over? 2Co + 3Cl2  2CoCl3 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). LR ER 57.92L Cl2 -19.41L Cl2 38.51LCl2 Left over:

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