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R2.1.5 Atom economy

Mike Sugiyama Jones (MSJ Chem)2:47 50.728 Aufrufe veröffentlicht Auf YouTube

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  1. this is msj CM in this video I'll be looking at atom economy here we have the equation for calculating percentage atom economy that appears in the IB data
  2. booklet so that's percentage atom economy equals the molar mass of the desired product divided by the molar mass of all reactants multiplied by
  3. 100 there are other versions of the equation and they are atom economy equals mass of desired product divided by mass of all reactants multiplied by
  4. 100 and mass of desired product divided by mass of all products multiplied by 100 to calculate the atom economy for a reaction either one of these equations
  5. can be used in this video I'll be using the top equation because this one appears in the IB data booklet next we look at an example ion
  6. is produced by the reduction of ion oxide in a blast furnace Cal the atom economy of the reaction here we have the balanced
  7. chemical equation for the reaction ion oxide reacts with carbon monoxide to form ion and carbon dioxide we'll start by finding the total
  8. of all the molar masses of the atoms in the reactants so we have two atoms of ion so that's 2 multiplied by the molar mass of ion which is 55.8 5 we have six
  9. atoms of oxygen so that's 6 multiplied by the mol mass of oxygen oygen which is 16. and we have three carbon atoms so that's 3 multiplied by the mol mass of
  10. carbon which is 12.01 when we add these together we get a total of 243.166
  11. us 11.7 next we'll use the equation to calculate the percentage atom economy
  12. the total of the molar masses of our desired product which was ion is 11.7 the total of the molar masses of the reactants was
  13. 243.166 which I rounded to two significant figures this percentage atom economy
  14. tells us that 54% by mass of the reactants do not end up in the desired product in other words they are wasted the higher the atom economy for a
  15. chemical reaction the less waste is produced and the more efficient the reaction is

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