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Unit: Stoichiometry

Day 5 - Notes. Unit: Stoichiometry. Excess reactant calculations. 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.

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Unit: Stoichiometry

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  1. Day 5 - Notes Unit: Stoichiometry Excess reactant calculations

  2. 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

  3. 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 =

  4. LR ER • 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 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 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 =

  5. LR ER • 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. K: 3.478x1023atomsCo U: ? L Cl2 3 mol Cl2 2 mol Co 22.4L Cl2 1 mol CoCl3 3.478x1023atomsCo 1 1 mol Co 6.02x1023atomsCo x x x = 19.41L Cl2 Reacts:

  6. LR ER • 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). 57.92L Cl2 -19.41L Cl2 38.51L Cl2 Left over:

  7. Questions? Complete WS 5 for HOMEWORK!

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