II. Stoichiometry in the Real World * Limiting Reagents More Stoichiometry! II. Stoichiometry in the Real World * Limiting Reagents

X A. Limiting Reactants Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly Limiting Reactant bread Excess Reactants peanut butter and jelly

X Jam Sandwich Equation 2 loaves bread + 1 jar of jam  20 sandwiches Mole Ratio 2 : 1 : 20 What would happen if we had 1 loaf of bread and 1 jar of jam? Bread = limiting reactant Jam = excess reactant (1/2 jar left) Produce 10 sandwiches

Limiting Reactant used up in a reaction determines the amount of product Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle

Example 1 2 Fe (s) + O2 (g)  2 FeO (s) Iron burns in air to form a solid black oxide (FeO). If you had 50.0 g of iron and 60.0 g of oxygen, what is the limiting and excess reactant? How much iron oxide could be produced? 2 Fe (s) + O2 (g)  2 FeO (s)

2 Fe (s) + O2 (g)  2 FeO (s) mass = 50 g mass = 50 g mass = 60 g

Example 2 Zn + 2HCl  ZnCl2 + H2 79.1 g mass = ? 0.90 L 2.5M 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many grams of zinc chloride would be produced? Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M mass = ?

Zn + 2 HCl  ZnCl2 + H2 mass = 79.1 g V = 0.90 L of 2.5M mass = ?

Example 2 Results Limiting reactant: HCl Excess reactant: Zn Product Formed: x g ZnCl2 left over zinc

Student Example CaCO3 + 2 NaCl  CaCl2 + Na2CO3 Sodium carbonate is needed in the manufacturing of glass, but very little occurs naturally. It can be made from the double replacement reaction between calcium carbonate and sodium chloride. If you had 5.00 g of each what is the limiting and excess reactant? How much sodium carbonate would be formed? CaCO3 + 2 NaCl  CaCl2 + Na2CO3

CaCO3 + 2 NaCl  CaCl2 + Na2CO3 mass = 5.0 g mass = 5.0 g mass = ?

Assignment Stoichiometry: Limiting Reagent Worksheet

B. Percent Yield measured in lab calculated on paper

Ivan Buz

When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

Example K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: sm mass = 45.8 g mass = ? actual: 46.3 g # moles = mm sm # moles = 0.662 mol mass = (# mol) (mm) # moles = 138.2 g/mol 45.8 g mass = (0.662 mol) (74.6 g/mol) # moles = 0.331 mol mass KCl = 49.4 g Theoretical Yield:

Example Continued Theoretical Yield = 49.4 g KCl Actual Yield  100 % % Yield = = 46.3 g 49.4 g  100 % = 93.7 %

Assignment Pg 295 # 2, 5, 7, 11, 12, 15, 18, 20, 25 - 30