Presentation on theme: "Chapter 11: Stoichiometry Sec. 11.3: Limiting Reactants."— Presentation transcript:
Chapter 11: Stoichiometry Sec. 11.3: Limiting Reactants
Objectives §Identify the limiting reactant in a chemical equation. §Identify the excess reactant and calculate the amount remaining after the reaction is complete. §Calculate the mass of a product when the amounts of more than one reactant are given.
Limiting Reactants If I want to make s’mores, does it matter how many marshmallows I have if I only have one piece of chocolate? No!! I will use up the chocolate & there will be marshmallows left over!
Limiting Reactants – p. 379, Fig. 4 §If I have 10 screwdrivers, 5 pliers, & 4 hammers, how many tool sets could I make? §Each tool set consists of 2 screwdrivers, a pliers and a hammer. §The hammers are used up. The number of hammers limits how many sets can be made. §There are leftover pliers and screwdrivers. There is an excess of these tools.
In Chemical Reactions: §A reaction will proceed until one reactant is used up. §The amount of product formed depends upon the reactant that is limited. When this reactant is used up, the reaction stops. §The limiting reactant is the reactant that limits the extent of the reaction and determines the amount of product formed.
In Chemical Reactions: §When the limiting reactant is used up, a portion of all of the other reactants will remain after the reaction stops. §These left-over reactants are called excess reactants.
Consider this reaction: 3H 2 + 3N 2 --> 2NH 3 + 2N 2 Visualize all the nitrogen and hydrogen atoms separating. The atoms are then available to reassemble into molecules (like the tools are assembled into kits.) Only two ammonia molecules can be assembled because there are 6 hydrogen atoms and each ammonia molecule requires 3! When the hydrogen is gone, some nitrogen atoms remain unreacted. Hydrogen is limiting and nitrogen is in excess.
Determining the limiting reactant & amount of product §If 200g of sulfur reacts with 100 g of chlorine, what mass of disulfur dichloride is produced? S 8 (l) + 4 Cl 2 (g) → 4S 2 Cl 2 (l) Remember: The amount of product depends on the reactant that is limited. Finding that is your first step!
S 8 (l) + 4 Cl 2 (g) → 4S 2 Cl 2 (l) §From the masses of reactants given, determine the number of moles of reactants l 100 g Cl 2 x 1 mol = 1.41 mol Cl 2 71 g l 200 g S 8 x 1 mol = mol S g
S 8 (l) + 4 Cl 2 (g) → 4S 2 Cl 2 (l) §Next, determine whether the 2 reactants are in the correct mole ratio. The equation indicates that the ratio of chlorine to sulfur is 4:1. Find the actual mole ratio by dividing the mole values you just calculated. l 1.41 mol Cl 2 = 1.81 mol Cl mol S 8 mol S 8
The Limiting Reactant l This means that 1.81 mol of chlorine is available for each mole of S 8. l The mole ratio says we need 4 chlorine for 1 sulfur. l There is not enough chlorine to use up all the sulfur so chlorine is the limiting reactant!!
Practice Problems §Identify the limiting reactant when 3.50 g of HCl reacts with 5.28 g of NaOH to produce NaCl and water. §Identify the limiting reactant when 1.22 g O 2 reacts with 1.05g H 2 to produce water.
Determining the amount of product §Since the limiting reactant determines the amount of product formed, the amount of product can be determined when the limiting reactant is known. §The moles of the limiting reactant will be our “known” in a standard stoichiometric calculation. The mass of the product is the unknown.
S 8 (l) + 4 Cl 2 (g) → 4S 2 Cl 2 (l) 1.41 mol ? Grams l 1.41 mol Cl 2 x 4 mol S 2 Cl 2 4 mol Cl 2 x g = 191 g S 2 Cl 2 1 mol Let’s use the example from before. We determined that Cl 2 was limiting and that there were 1.41 mol of it.
Practice Problems §Determine the mass of tetraphosphorus decoxide formed if 25.0 g of phosphorus (P 4 ) and 50.0 g oxygen are combined. P 4 + 5O 2 --> P 4 O 10 §In photosynthesis, 6CO 2 + 6H 2 O --> C 6 H 12 O 6 + 6O 2. If a plant has 88.0 g of CO 2 and 64.0 g H 2 O available, what mass of glucose will be produced?
Excess Reactants §Once the limiting reactant has been determined, it is also possible to use stoichiometry to find out how much of the excess reactant is used or how much is leftover. §Recall this reaction: S 8 (l) + 4 Cl 2 (g) → 4S 2 Cl 2 (l) We determined the limiting reactant is chlorine, with the amount of 1.41 moles.
S 8 (l) + 4 Cl 2 (g) → 4S 2 Cl 2 (l) With the known of 1.41 mol chlorine, we can determine the no. of moles and grams of sulfur that will be used: 1.41 mol Cl 2 x 1 mol S 8 = mol S 8 4 mol Cl mol S 8 x g = 90.7 g S 8 USED 1 mol Subtract to get the amount that is left over: 200 g (the original amount of S 8 ) – 90.7 g = g LEFTOVER
Practice Problems 1. What mass of SO 3 is produced from the reaction of 10.5 g of SO 2 and 4.32 g of O 2 ? How much of the excess reactant is left when the reaction is complete? 2. If 8.0 g of Cr is heated with 16.7 g of Cl 2, what mass of CrCl 3 will be produced? What is the excess reactant and how much of it is used in the reaction?