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Starter. A sandwich consists of two slices of bread, 3 slices of meat, and one slice of cheese. For each of the following amounts, determine the number of sandwiches that can be made and what is left over: 6 bread, 10 meat, 4 cheese slices 10 bread, 6 meat, 8 cheese slices

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  1. Starter A sandwich consists of two slices of bread, 3 slices of meat, and one slice of cheese. For each of the following amounts, determine the number of sandwiches that can be made and what is left over: 6 bread, 10 meat, 4 cheese slices 10 bread, 6 meat, 8 cheese slices 25 bread, 15 meat, 12 cheese slices

  2. Starter • 6 bread, 10 meat, 4 cheese slices • 3 sandwiches • 0 bread, 1 meat, 1 cheese • 10 bread, 6 meat, 8 cheese slices • 2 sandwiches • 6 bread, 0 meat, 6 cheese • 25 bread, 15 meat, 12 cheese slices • 5 sandwiches • 15 bread, 0 meat, 7 cheese

  3. Ch. 9 Stoichiometry 9.3 Limiting Reactant and Percent Yield

  4. Why is there a limiting reactant? • a reaction rarely has exactly the right amount of each reactant • usually have some left over of one of them • limiting reactant • reactant that limits the amount of product created • completely used up • excess reactant • reactant not completely used up

  5. When do you have to find a LR? whenever two amounts of reactants are given when only one amount of reactant is given, then the other is assumed to be in excess

  6. Finding Limiting Reactant • How much do you have to start? • Find the number of moles of both reactants • Figure out how much you need of B if you use up all of A • Convert moles A to moles B using mole ratio • You may start with either one • Determine whether you will have enough • If you don’t have enough of B, then B is LR • If you do have enough of B, then A is LR

  7. Example 1 The reaction begins with 2.51 g of HF and 4.56 g of SiO2. What is the limiting reactant and the excess reactant? Write the balanced chemical equation SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)

  8. Example 1 • Find the number of moles available of each reactant: HAVE

  9. Example 1 • If we use up all of the HF, how much SiO2 will we need to go with it? • Do we have enough SiO2? • 0.0759 mol available > 0.0313 mol needed • YES- there will be some left over • Limiting Reactant : HF NEED

  10. How much ER is left over? • Find out the number of moles of ER used up in the reaction • convert moles of LR to moles of ER using mole ratio • Subtract that amount from the moles you started with

  11. Example 1 • How many grams of SiO2 will be left over? • Find the moles of SiO2 used up. • Subtract that from the moles started with • 0.0759 mol available - 0.0313 mol needed = .0446 moles SiO2 left over • Convert moles to grams using molar mass

  12. How much of the product can be formed? Start conversion with moles of limiting reactant. Convert to moles of product using mole ratio Convert to grams if requested using molar mass.

  13. Example 1 • How many grams of water could be formed? • Convert moles of HF to moles of water • Convert moles to grams using molar mass.

  14. Why percent yield? • Usually, not all the product possible is actually formed. • because of error in lab procedure • theoretical yield • maximum amount of product possible • actual yield • the measured amount formed in reaction

  15. Example 1 • If the reaction actually produced 1.02 grams of water, what is the percent yield?

  16. Example 2 A reaction was done with 36.8 g C6H6 and 41.0 g of O2. Write the balanced chemical equation 2C6H6 + 15O2 12CO2 + 6H2O

  17. Example 2 • What is the limiting reactant? • Find the number of moles of each • Convert one to the other • Compare to amount available • don’t have enough so O2 is LR and C6H6 is ER HAVE

  18. Example 2 • Find the amount of ER left over • Find the moles ER used • Subtract from moles starting • 0.471 moles available – 0.171 moles used = 0.300 moles left over

  19. Example 2 • Find the percent yield if 32.0 g of CO2 was created. • Find theoretical yield of CO2 • Calculate percent yield

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