To identify which of the reactant is the limiting reagent and which one is the excess, you have to assume that one substance is limited, then compute for the required amount of the other reactant. Let's us use again the equation N2 3H2 2NH3 10 g 10 g If you have 10 grams each of Na and Hz, which one is the limiting reagent and which one is the excess reagent? To answer this, you have to assume on reactant to be a limiting reagent, then compute for the required amount of the other reactant to completely consume the other. Let us assume that N2 is the limiting reagent. How much Hz do you need to completely use up all the N2? Convert Mass A Convert Mole B to Mole A to Mass B Mole Ratio 10 g N2 (t Nth Nt 0.36 = mol N2 (3 Nth Ht1.1 = mol H2 (ts,t g Ht = 2.2 g H2 (t Nt HE (test gh As you can see from the calculation, 10 grams of N2 will only require 2.2 grams of H2. Since you have 10 grams of Hz, it means that it is in excess while Nz is in limited quantity. But what if you assumed that H2 was the limiting reagent? How much N2 do you need to consume all 10 grams of Ha? Convert Mass A Convert Mole B to Mole A to Mass B Mole Ratio 10 g H2 (4 5.0 mol Hz ft Nth Nt1.7 = mol N2 (ISsig 48 g N2 (tsuglit 3) wth HE (t Nth-Nr If all 10 grams of Hz get used up in the reaction, you will be needing 48 grams of N2. Since you only have 10 grams of N2, it means that its quantity is limited and, therefore, H2 is in excess.Activity 3: Am I too much for you? Objectives: Follow the steps in solving for the limiting and excess reagents carefully; and Identify the limiting and excess reagents. Procedure: Solve the following problems to identify the limiting and excess reagent using a single thread of solution, just like in the example . Report your answer in the correct number of significant figures 1. In the reaction Mg(OH)2 + 2HCI - MgCl, + 2H2 O, you were given 15 grams of Mg(OH)2 and 25 grams of HCI. Which of the two reactants is the limiting reagent? How much MgCl2 will be produced in the reaction? (Mg(OH)2 = 58.32 g/mol, HCI = 36.46 g/mol, MgCl2 = 95.32 g/mol) 2. Nitric oxide (NO) is a substance that can improve heart health and enhance your performance during exercise or workout. It can be formed through the reaction: NH3 + 02 -> NO + H2O If 25 g NH3 is reacted with 20 g O2, which of the two is the limiting reagent? How much NO will be formed in the reaction? (NH3 = 17.03 g/mol, Oz = 32.00 g/mol, NO = 30.01 g/mol) 3. Consider the reaction MnO2 + 4HCI -> MnCl2 + Cl2 + 2H2O. If the both the reactants have a mass of 100 grams, which of the two will be used up first? How many grams of MnCl will be produced after the reaction? (MnO2 = 86.94 g/mol, HCI = 36.46 g/mol, MnCl2 = 125.8 g/mol)If you are given a specific mass of substance, how do you know how much of the other reactants are you going to use? Can you also know the amount of product that you are going to form? To solve this, you need to follow the steps in solving stoichiometric problems: Balance Convert Convert Mole A Convert the Mass A to Mole B using Mole B chemical to Mole the Molar Ratio to Mass equation A B STEP 1 STEP 2 STEP 3 STEP 4 For example, in the reaction Mg + HNO; > H2 + Mg(NO;)2, you were given 10 grams of Mg, how much HNO, are going to use and how much H2 and Mg(NO,), will be formed? STEP 1: Balance the chemical equation. The balanced chemical equation is Mg + 2HNO, > H2 + Mg(NO,)2 STEP 2: Convert Mass of Mg to Mole 10 g Mg( ' with Mg ) = 0.41 mol Mg 12/s arg Mg STEP 3: Convert Mole of Mg to the Mole of the desired substance using Mole Ratio (Based from the balanced chemical equation) 0.41 mol Ma(* with HN13) = 0.82 mol HNO, ENting 0.41 mol Mg('Nth H) = 0.41 mol Hz tNUMB 0.41 mot Mg(t Nth Mg N13 1) thing 0.41 mol Mg(NO3)2 STEP 4: Convert Mole of the desired substance to Mass 0.82 mol HNO3 (63s.t g HNt3) = 52 g HNO3 t NUHNE3 0.41 mol Hz(ts,t g Ht) = 0.83 g H2 t Nemfit 0.41 mol Mg(NO3)2(12/353 g Mg NE30) 61 g Mg(NO3)2Activity 2: Cook like a Chemist Imagine you are a chemist. How do you know how much reactants you are going to use and how much products are going to be formed? Try solving these problems to practice your skills. Objectives: Solve the following stoichiometric problems correctly; and Follow the steps in solving stoichiometric problems carefully Procedure: Follow the four steps in solving stoichiometric problems to solve for what is being required. . Write you final answer in the correct number of significant figures. 1. When baking soda (NaHCO,) is heated, carbon dioxide (CO2) is released. This makes bread, cookies and other pastries to rise. If 42.0 g of NaHCO, was used, how much CO2 was released? (NaHCO, = 84.00 g/mol, CO2 = 44.01, NazCO, = 105.99 g/mol, HO = 18.02 g/mol) 2NaHCO, -> NazCO3 + HO + CO2 2. How much potassium iodide (166.00 g/mol) is required to get 1.78 g of mercury(II) iodide precipitate (454.39 g/mol). The unbalanced chemical equation is: KI + Hg(NO,)2 -> Hglz + KNO, 3. Titanium (IV) oxide (TiOz) is a pigment used for white paint. It is formed from the reaction of the mineral ilmenite (FeTIO,) with sulfuric acid. How much TiOz will be produced if 1000.0 g of FeTio, is reacted with sufficient amount of sulfuric acid? (FeTio, = 151.71 g/mol, TiOz = 79.87 g/mol) FeTiO; + H,SO. > Tio, + FeSO, + HO There are times when the ingredients that we use when cooking dictates how much food we are going to make. In the case of our chicken adobo, if you have 10 pounds of chicken but you only have one onion, you will be forced to cook only one pound of chicken to avoid making your dish go bland. In the same manner, certain reactants with limited amount dictate the amount of product we are going to form. This is where the concept of limiting reagent and excess reagent comes in. A limiting reagent is the substance in a chemical reaction that controls or limits the maximum amount of product formed while an excess reagent is the reactant present in quantities greater than necessary to react with the quantity of the limiting reagent. In the example I gave earlier, the limiting reagent (ingredient) is the onion while the excess reagent (ingredient) are the chicken