STOICHIOMETRY & LIMITING REACTANTS UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS

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1 STOICHIOMETRY & LIMITING REACTANTS UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS If the number of atoms is conserved in a chemical reaction, the mass must also be conserved as expected from the Law of Conservation of Mass. In the equation for the formation of water H (g) + O (g) H O(l) molecules of hydrogen and 1 molecule of oxygen combine to form molecules of water. We could also say that moles of hydrogen react with 1 mole of oxygen to form moles of water. Using the number of grams in a mole of each substance, the mass relationships in the table can be determined. The ratio of moles of hydrogen to moles of oxygen in forming water will be :1. If 10 moles of hydrogen are available, 5 moles of oxygen are required. H (g) + O (g) H O(l) molecules 1 molecule molecules mol 1 mol mol moles x (.0g/mol) 1 mole x (3.00g/mol) moles x (18.0g/mol) 4.04 g 3.00 g g Solving problems involving the masses of products and or reactants is conveniently accomplished by dimensional analysis. All numerical problems involving chemical reactions begin with a balanced equation. Example Find the mass of water produced when 10.0 grams hydrogen react with plenty of oxygen. H (g) + O (g) H O(l) Step 1: We know the amount of oxygen is in excess, so we will focus on hydrogen. Find the moles of hydrogen represented by 10.0 g by using the molar mass of H. 1mol H 10.0 g H = 4.95 mol H.0 g H Step : Find the number of moles of H O produced by 4.95 mole H. From the balanced equation, we know that for every mol H we produce mol H O. mol H O 4.95 mol H = 4.95 mol H O mol H Step 3: Find the mass of H O that contains 4.95 mol H O by using the molar mass of water g HO 4.95 mol H O = 89.1 g HO 1mol HO Most chemistry students find it is more convenient to set up all three steps in one problem. Make sure that all labels cancel except the g H O, an appropriate unit to express the mass of H O as required in the problem. 1mol H mol H O 18.0 g H O 10.0 g H = 89.1 g H O.0 g H mol H 1mol H O Molar mass Coefficients Molar Mass of H in equation of H O

2 Understanding Limiting Reactants In Unit 7, Section B.9 of your ChemCom textbook you will find a conceptual atom-counting method for finding the limiting reactant of an equation. If a combustion problem states that excess oxygen is available, we need not concern ourselves with the oxygen-to-water ratio. The mass of water produced is predicted from (and limited by) the mass of hydrogen available. Of course, in the laboratory we often deal with specified masses of each reactant, but that requires an enhanced problem-solving method. Suppose a family wants to make several chile rellenos casseroles to serve at a neighborhood party. The recipe lists required ingredients, which are itemized in the left column of the table in Figure 3. The right column represents a survey of pantry and refrigerator contents. Required Ingredients Available Ingredients Possible Casseroles 1 7-oz can whole green 7-oz cans whole green chiles chiles 1 pound Monterrey Jack 3 pounds Monterrey Jack 3 cheese Cheese 1 pound cheddar cheese 3 pounds cheddar cheese 3 3 eggs 1 dozen eggs 4 3 Tablespoons flour 5 pounds flour Many 5 oz canned evaporated milk 4 cans evaporated milk 4 Figure 3: Required Ingredients Table How many casseroles can be made? Although four casseroles can be made from the available eggs or milk, there are only enough cans of green chiles for two casseroles. In other words, the number of cans of green chiles can be called the limiting factor. After the two casseroles are prepared, cheese, eggs, flour, and milk will remain, but all the green chiles will be used. Therefore, no more casseroles can be made. In this example, the green chiles are the limiting reactant. The limiting reactant is the reactant that is consumed first and limits the amount of product that can be made. The same principle applies in determining the quantity of product that can be produced in a chemical reaction. Let s take another look at the reaction of hydrogen and oxygen to produce water, then consider what would happen if.00 mol hydrogen and.00 mol oxygen were available. How many moles of water can be produced? What is the limiting reactant? Which reactant will be in excess and by how much? H (g) + O (g) H O(l) The balanced chemical equation states that.00 mol hydrogen react with 1.00 mol oxygen. When the reaction is complete,.00 mol water are produced and 1.00 mol oxygen remains unreacted. This problem is easy to solve by inspection. A more systematic way to solve the problem is to create an SRF table, as shown in Figure 4. The table is composed of the following lines: Line 1: The balanced chemical equation is listed. Line : The Starting number of moles of each substance is listed. This would be what is available, the same as the ingredients for the casseroles. Line 3: The Reacting ratio determined from the coefficients in the balanced equation is multiplied by x, the basic amount of moles that will react. The reactants are being consumed so a minus sign is placed in front of them. The products are increasing so a plus sign is placed in front of them. Line 4: This line contains what will be left and what will be formed when the reaction is complete. The values of this line are obtained by adding lines and 3. This line will provide all the answers, in Final moles, to the problems.

3 Line 1 H + O H O Line Starting moles 0 Line 3 Reacting moles -x -1x +x Line 4 Final moles -x -1x 0+x Figure 4 Sample SRF Table When the reaction is completed, either the hydrogen will be consumed or the oxygen will be consumed. That means that either x = 0 (hydrogen) or - 1x = 0 (oxygen) If we solve for x in both equations the smallest x value will be the limiting reactant. The larger value will represent an amount in excess of what is possible with the given ingredients. So for this example, x = 1 (hydrogen) or x = (oxygen). Since x = 1 is smallest, it is the value we choose. This step also identifies the substance that will run out first, the limiting reactant. In this case the hydrogen is the limiting reactant. Using the value of x = 1, we can determine the final amounts. The amount of product is x = (1) = moles of H O. The reactant in excess is O and it is excess by 1x = 1(1) = 1 mole of O. This method of working the problem gives the same result as the visual inspection hydrogen is the limiting reactant, and.00 mol H O are produced. The advantage of learning this method is that it works even when the coefficients become difficult to use with the visual method. Example Aluminum chloride, AlCl 3, has many uses including in deodorants and antiperspirants. It is synthesized from aluminum and chlorine. What mass of AlCl 3 can be produced if 100 g of each reactant are available? What is the limiting reactant? How many grams of the excess reactant remain? Step 1: Write the balanced equation. Al(s) + 3 Cl (g) AlCl 3 (s) Step : Set up the SRF Table. Since only moles can go into the table, the grams of each reactant will first need to be converted to moles. See Figure 4. 1 mol Al 100 g Al = 3.70 moles Al 7.0 g Al 100 g Cl 1 mol Cl 71.0 gcl = moles Cl Al(s) + 3 Cl (g) AlCl 3 (s) Starting moles Reacting moles -x -3x +x Final moles x x x Step 3: Set the Final moles equal to zero and solve for x. Choose the smallest value of x x = 0 or x = 0 x = 1.85 moles or x = 0.47 moles

4 Since x = 0.47 mol is smaller, Cl will be consumed first; it is the limiting reactant. Step 4: Determine the amount of product formed and the amount of Al remaining by substituting x = 0.47 into the corresponding equations on the Final moles line. Amount of product (AlCl 3 ): x = (0.47 mole) = 0.94 moles AlCl 3 Amount of Al remaining: 3.70 (0.47 mole) =.76 moles of Al Step 5: Convert the moles back to grams. 134 grams 0.94 moles AlCl3 = 16 grams AlCl3 formed 1 mole.76 moles Al 7.0 grams 1 mole = 74.5 grams Al left Practice Exercises 1. Acetylene burns in air to form carbon dioxide and water: C H (g) + 5 O (g) 4 CO (g) + H O(l) How many moles of CO are formed from 5.0 moles C H?. If insufficient oxygen is available, carbon monoxide can be a product of the combustion of butane: C 4 H 10 (l) + 9 O (g) 8 CO(g) + 10 H O(l). What mass of CO could be produced from 5.0 g butane? g NaNH is required for an experiment. Using the following reaction, what mass of sodium is required for reaction. Na(s) + NH 3 (g) NaNH (s) + H (g)?

5 4. Heating CaCO 3 yields CaO and CO. Write the balanced equation. Calculate the mass of CaCO 3 consumed when 4.65 g of CaO forms. 5. In the synthesis of sodium amide (NaNH ), what is the maximum mass of NaNH possible if 50.0 g of Na and 50.0 g NH 3 were used? Na(l) + NH 3 (g) NaNH (s) + H (g) 6. The fuel methanol, CH 3 OH, can be made directly from carbon monoxide (CO) and hydrogen (H ). a. Write a balanced equation for the reaction. b. Calculate the maximum mass of methanol if one starts with 5.75 g CO and 10.0 g H. c. Which reactant is the limiting reactant? 7. Aspirin (C 9 H 8 O 4 ) is synthesized in the laboratory from salicylic acid (C 7 H 6 O 3 ) and acetic anhydride (C 4 H 6 O 3 ): C 7 H 6 O 3 (s) + C 4 H 6 O 3 (l) C 9 H 8 O 4 (s) + CH 3 COOH(l) a. What is the theoretical yield of aspirin if you started with 15.0 g salicylic acid and 15.0 g acetic anhydride? b. Which reactant is the limiting reactant?

6 HW_STOICH KEY 1) 50mol CO ) molar mass of butane = 58.14g/mol => mol butane => 0.344mol CO => 9.6g CO 3) molar mass of NaNH = 39.0g/mol => mol NaNH => mol Na => 8.84g Na 4) CaCO 3 à CaO + CO a. 4.65gCaO => 0.089molCaO => 0.089mol CaCO3 => g/mol x 0.089mol = 8.30gCaCO 3 5) 84.7g NaNH 6) CO(g) + H (g) à CH 3 OH(l) 0.05mol or 6.56g 7) 0.109mol = 19.5g