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Limiting Reagents and Percent Yield

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Presentation on theme: "Limiting Reagents and Percent Yield"— Presentation transcript:

1 Limiting Reagents and Percent Yield

2 Which of the Items Limits the Number of Burgers That Can Be Made?

3 Limiting Reagents

4 Vocabulary Limiting Reagent
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. In other words– the stuff you run out of first Excess Reagent The reactant that is not completely used up in a reaction.

5 Steps Write and balance the reaction equation
Convert given information to moles Use the mole ratio (from the balanced reaction) to determine which reactant will run out first. Once you determine the limiting reagent you can use that information to determine the maximum amount of product that can be made.

6 Example 79.1 g of zinc react with 63.5 g of HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Write the balanced reaction. Zn HCl > H ZnCl2 2. Covert the given information to moles 79.1 g Zn mol Zn = mol Zn g Zn 63.5 g HCl mol HCl = mol HCl g HCl 3. Use the mole ratio to determine which reactant will run out first. 1.21 mol Zn mol HCl = mol HCl 1 mol Zn For 1.21 mol of Zn you would need 2.42 mol of HCl, but you only have 1.74 mol HCl, that means the HCl will run out first and be the limiting reagent

7 Example Continued How many liters of hydrogen are formed at STP? Since HCl is the limiting reagent we use the amount of moles that it has available to find the amount of H2 that is formed mol HCl 1 mol H2 = mol H2 2 mol HCl Now we need to use Mole Island to convert from moles of H2 gas to liters of H2 gas mol H L = 19.5 L of H2 gas at STP 1 mol

8 Percent Yield

9 Equation Percent Yield = actual yield X 100 theoretical yield

10 Example When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. The actual amount produced in 46.3 g of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 Solve for the theoretical yield of KCl by setting up a stoichiometric conversion. 45.8 g of K2CO mole K2CO mol KCl g KCl g K2CO mol K2CO mol KCl = 49.4 g KCl

11 Example Continued 2. Solve for percent yield. Actual yield: 46.3g KCl Theoretical yield: 49.4g KCl Percent Yield = 46.3/49.4 X 100 = 93.7 %

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