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 firstExcess ReagentThe reactant that is not completely used up in a reaction.
5 Steps Write and balance the reaction equation Convert given information to molesUse 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 Example79.1 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 ZnCl22. Covert the given information to moles79.1 g Zn mol Zn = mol Zng Zn63.5 g HCl mol HCl = mol HClg HCl3. Use the mole ratio to determine which reactant will run out first.1.21 mol Zn mol HCl = mol HCl1 mol ZnFor 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 ContinuedHow 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
9 EquationPercent Yield = actual yield X 100 theoretical yield
10 ExampleWhen 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 + CO2Solve for the theoretical yield of KCl by setting up a stoichiometric conversion.45.8 g of K2CO mole K2CO mol KCl g KClg K2CO mol K2CO mol KCl= 49.4 g KCl
11 Example Continued2. Solve for percent yield. Actual yield: 46.3g KCl Theoretical yield: 49.4g KCl Percent Yield = 46.3/49.4 X 100 = 93.7 %