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Limiting Reactants

Limiting Reactants. 1/15/08. Some Definitions. Limiting Reactant – The reactant that limits the ability/amounts of other reactants to combine. Also limits the amount of product that can form in a chemical reaction. It can also be called the limiting reagent.

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Limiting Reactants

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  1. Limiting Reactants 1/15/08

  2. Some Definitions • Limiting Reactant – The reactant that limits the ability/amounts of other reactants to combine. Also limits the amount of product that can form in a chemical reaction. • It can also be called the limiting reagent. • The substance that is not used completely is the excess reactant.

  3. Analogy – • If you have 400 people wanting to fly to Hawaii and 350 seats on the plane, how many people can go on the flight? • So what is the limiting reagent? • What is the excess reactant? 350 The seats The people

  4. How to find the limiting reactant • Use the given amount of a reactant to determine the needed amount of the other reactant. • Compare the calculated (needed) amount to the actual given amount. • If you have more than you need the substance is in excess.

  5. Now apply it to Chemistry • If 2.0 mol HF are exposed to 4.5 mol of SiO2, which is the limiting reactant? • Since you need 0.5 mol of SiO2 and have 4.5mol then it is in excess. • So the answer is…..HF SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) 2.0 mol HF x 1mol SiO2 = 0.50 mol SiO2 4mol HF

  6. How ‘bout one more? • Which is the limiting reactant when 0.750 mol of N2H4 is mixed with 0.500 mol of H2O2? • How much of the excess reactant, in moles remains unchanged? N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(g) H2O2 0.750 mol N2H4 x 2 mol H2O2 = 1.50 mol H2O2 1 mol N2H4 0.500 mol N2H4 are left 0.500 mol H2O2 x 1 mol N2H4 = 0.250 mol N2H4 2 mol H2O2

  7. Percent Yield • Theoretical Yield – The maximum amount of product that can be produced from a given amount of reactant. • Actual Yield – The measured amount of a product obtained from a reaction of that product. • Percent Yield – The ratio of the actual yield to the theoretical yield, multiplied by 100.

  8. Let’s Practice • C6H6(l) + Cl2(g) → C6H5Cl(s) + HCl(g) • When 36.8g of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8g. What is the percent yield of C6H5Cl?

  9. The Plan • Find the Theoretical yield with a mass-mass calculation • Use the Theoretical yield and Actual yield to calculate the Percent yield. 36.8 g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.5 g C6H5Cl = 53.0 g C6H5Cl 78.06 g C6H6 1 mol C6H6 1 mol C6H5Cl 38.8 g x 100 = 73.2 % 53.0 g

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