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

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Presentation on theme: "Limiting Reactants and Percent Yields Objectives: 1. Define limiting reactant. 2. Explain how the quantities of products are determined in a chemical reaction."— Presentation transcript:

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 N H 2  2NH 3 If 10 moles of N 2 react with 10 moles of H 2, how many moles of NH 3 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 N 2 = 10 10/3 H 2 = 3.33 – the reactant with the smallest quantity is the limiting reactant 2nd step - solve the equation using mol - mol 10 mol H 2 x 2 mol NH 3 /3 mol H 2 = 6.7 mol NH 3 At the end of the reaction there is an excess of N 2 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 N H 2  2NH 3 10 mol H 2 x 2 mol NH 3 /3 mol H 2 = 6.7 mol NH 3 In the above problem 112.9g of NH 3 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 NH 3 x 17g NH 3 / 1mol NH 3 = 113.9g NH 3 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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