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Limiting Reactants

Limiting Reactants. Stochiometry Lesson 2 Chemistry. Limiting Reactants. Limiting reactants limit the extent of the reaction and determine the amount of the product. Excess reactants are the left-over reactant at the end of the reaction. Calculating the Product when the Reactant is Limited.

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Limiting Reactants

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  1. Limiting Reactants Stochiometry Lesson 2 Chemistry

  2. Limiting Reactants Limiting reactants limit the extent of the reaction and determine the amount of the product. Excess reactants are the left-over reactant at the end of the reaction.

  3. Calculating the Product when the Reactant is Limited • To determine the limiting reactant, convert the mass of both reactants to moles. • Determine if the reactants are in the correct mole ratio. • Divide the available moles of one reactant by the available moles of the second reactant. • Compare the mole ratio of the available to the mole ratio in the balanced equation. • The limiting reaction will be the reactant that has less moles than expected from the balanced equation. • The limiting reactant will be used to determine the number of moles of product that can be produced.

  4. Example Problem The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflates an automobile airbag. 6 Na (s) + Fe2O3 (s)  3 Na2O (s) + 2 Fe (s) If 100 g Na and 100 g Fe2O3 are used in this reaction, determine the limiting reactant c. mass of the iron produced reactant in excess d. mass of excess reactant left over

  5. Limiting reactant and excess reaction 6 Na (s) + Fe2O3 (s)  3 Na2O (s) + 2 Fe (s) 100 g Na/ 22.99 g Na/mol = 4.35 mol Na 100 g Fe2O3 / 159.69 g/mol = 0.636 mol Fe2O3 4.35 mol Na/0.636 mol Fe2O3 = 6.84 0.636 mol Fe2O3 / 0.636 mol Fe2O3 = 1 Compare ratio to balance equation: 6 Na : 6.84 Na 1 mol Fe2O3: 0.636 mol Fe2O3 Fe2O3 is the limiting and Na is the excess

  6. Mass of the solid iron produced 6 Na (s) + Fe2O3 (s)  3 Na2O (s) + 2 Fe (s) 0.636 mol Fe2O3 (limiting) x 2 Fe/1 Fe2O3 = 1.27 mol Fe x 55.84 g/mol Fe = 70.9 g Mass of excess reactant that remains 0.636 mol Fe2O3 (limiting) x 6 mol Na/1 mol Fe2O3 = 3.82 mol Na x 22.99 g/mol Na = 87.82 g Na used 100 g – 87.82 g used = 12.189 g Na extra

  7. Percent Yield • The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. • A chemical reaction RARELY produces the theoretical yield. • The actual yield is the amount of product actually produced when the chemical reaction is carried out in an experiment. • Percent yield of the product is the ratio of the actual yield to the theoretical yield expressed as a percent. • Percent yield = (actual yield ÷ theoretical yield) x 100 %

  8. Example Problem When potassium chromate (K2CrO4) is added to a solution containing 0.500 g silver nitrate (AgNO3), solid silver chromate (Ag2CrO4) is formed. Determine the theoretical yield of the Ag2CrO4. If 0.455 g Ag2CrO4 is obtained, calculate the percent yield.

  9. 2 AgNO3 + K2CrO4  Ag2CrO4+ 2 KNO3 0.500 g AgNO3 ÷ 170 g/mol = 0.00294 mol 0.00294 mol AgNO3 x 1 Ag2CrO4/2 mol AgNO3 = 0.00147 mol Ag2CrO4 x 332 g/mol = 0.488 g Ag2CrO4 Percent yield = 0.455 g (actual) ÷ 0.488 g (theoret.) = 0.932 x 100% = 93.2 %

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