Presentation on theme: "II. Stoichiometry in the Real World * Limiting Reagents More Stoichiometry!"— Presentation transcript:
II. Stoichiometry in the Real World * Limiting Reagents More Stoichiometry!
A. Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly X
Jam Sandwich Equation 2 loaves bread + 1 jar of jam 20 sandwiches b Mole Ratio2 : 1 : 20 b 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 X
b Limiting Reactant used up in a reaction determines the amount of product b Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle
b 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? Example 1 2 Fe (s) + O 2 (g) 2 FeO (s)
mass = 50 g mass = 60 g mass = ?
b 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 ZnCl 2 + H g mass = ? 0.90 L 2.5M Example 2
Zn + 2 HCl ZnCl 2 + H 2 mass = 79.1 g mass = ? V = 0.90 L of 2.5M
Example 2 Results Limiting reactant: HCl Excess reactant: Zn Product Formed: x g ZnCl 2 left over zinc
b 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? Student Example CaCO NaCl CaCl 2 + Na 2 CO 3
B. Percent Yield calculated on paper measured in lab
b When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl 2KCl + H 2 O + CO g? g actual: 46.3 g
ExampleExample K 2 CO 3 + 2HCl 2KCl + H 2 O + CO 2 mass = 45.8 g mass = ? actual: 46.3 g Theoretical Yield: # moles = mm sm # moles = g/mol 45.8 g # moles = mol # moles = mol mass = (# mol) (mm) mass = (0.662 mol) (74.6 g/mol) mass KCl = 49.4 g
Example Continued Theoretical Yield = 49.4 g KCl = 46.3 g 49.4 g 100 % = 93.7 % % Yield = Actual Yield Theoretical Yield 100 %