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Limiting Reactants and Percent Yields

Limiting Reactants and Percent Yields. Objectives: 1. Define limiting reactant. 2. Explain how the quantities of products are determined in a chemical reaction. 3. Explain how to calculate the percent yield of a reaction. Limiting Reactant:

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Limiting Reactants and Percent Yields

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  1. Limiting Reactants and Percent Yields Objectives: • 1. Define limiting reactant. • 2. Explain how the quantities of products are determined in a chemical reaction. • 3. Explain how to calculate the percent yield of a reaction.

  2. Limiting Reactant: • The limiting reactant in a chemical equation is defined as the reactant that limits the amount of product that can be produced. The unused reactant is left in excess after the reaction.

  3. Determining limiting reactants using ratios of coefficients Consider the following reaction N2+ 3 H2g 2NH3 • If 10 moles of N2 react with 10 moles of H2 , how many moles of NH3 will be produced? What is the limiting reactant? • 1st step is to find the limiting reactant. • divide the moles given by the coefficient listed in the formula • 10/1 N2 = 10 • 10/3 H2 = 3.33 • the reactant with the smallest quantity is the limiting reactant • 2nd step - solve the equation using mol - mol • 10 mol H2 x 2 mol NH3 /3 mol H2 = 6.7 mol NH3 • At the end of the reaction there is an excess of N2 remaining

  4. Percent Yield Percent Yield: • The experimental (actual) yield or amount of product obtained in a chemical reaction rarely matches the theoretical (predicted ) yield calculated through stoichiometric calculations. This could be due to a number of different situations including: • testing conditions • side reactions • errors in recovering the products

  5. Calculation for percent yield • Percent Yield = (actual yield /theoretical yield) x 100% Consider the following reaction N2+ 3 H2g 2NH3 10 mol H2 x 2 mol NH3 /3 mol H2 = 6.7 mol NH3 In the above problem 112.9g of NH3 are obtained from the reaction. What is the percent yield from the experiment? • 1st step - find the theoretical yield • mol - mass problem • 6.7 mol NH3 x 17g NH3 / 1mol NH3 = 113.9g NH3 • 2nd step - plug into formula • (112.9g / 113g) x 100% = 99% • If done correctly the value should ALWAYS be less than 100%

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