Gen. Chem1_Q1 Lesson 9_ Reaction Stoichiometry.pdf

Reaction Stoichiometry Cathyrine Canete | RSHS-SHS III 1. Determine mass relationship in a chemical reaction (STEM_GC11MRIg-h-42). 2. Explain the concept of limiting reagent in a chemical reaction; identify the excess reagent(s) (STEM_GC11MRIg-h-40); 3. Calculate percent yield and theoretical yie...

Reaction Stoichiometry Cathyrine Canete | RSHS-SHS III 1. Determine mass relationship in a chemical reaction (STEM_GC11MRIg-h-42). 2. Explain the concept of limiting reagent in a chemical reaction; identify the excess reagent(s) (STEM_GC11MRIg-h-40); 3. Calculate percent yield and theoretical yield of the (STEM_GC11MRIg-h-39);  STOICHIOMETRY is the quantitative relationship in Chemical Reaction which are illustrated in Chemical Equation which maybe used in solving problems such as Percentage mass-mole mole-mass mass-particles yield  Lesson 1 Limiting and Excess Reagent Excess Baggage Directions: The table on the next slide shows the primary materials needed to produce the items in the first column. With the given available materials, identify how many items are produced, which materials are in excess and how many are left of the available materials.  Item Required Available materials Produced Excess set materials Bicycle 1 bike frame, 2 tires 68 bike frames, 117 2 pedals, 1 crank tires, 250 pedals, 72 arm, 1 brake set crank arm, 93 brake set Banana 2 bananas, 1 236 bananas, 150 Cue barbecue stick, ½ barbecue stick, 25 cup brown sugar cups brown sugar Cheese 1 burger bun, 1 beef 324 burger buns, 12 burger patty, 1 slice of dozens beef patties, cheese 261 slices of cheese Table 1 table top, 4 legs, 8 20 table tops, 50 nuts, 8 screws legs, 50 nuts, 50 screws Milk tea 1 bag of black tea, 15 tea bags, 3 liters 250 ml water, 1/8 water, 2 cups milk, cup milk, 2 tbsp 30 tbsp sugar, 4 sugar, ¼ cup tapioca cups tapioca pearls pearls Reaction Stoichiometry - is the study of a quantitative relationship among reactants and products in a chemical reactions. Limiting reagent - the reagent that has completely reacted and used up in a reaction Excess reagent - is the reactant that is present in quantity higher than what is required to react with the limiting reagent  Let’s use that context to the balanced chemical 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 equation below: 3𝐻2(𝑔) + 𝑁2 (𝑔) → 2𝑁𝐻3 (𝑔) a. If 6.60 moles H2 are made to react with 4.42 moles N2, what is the limiting reagent? How many moles NH3 will be produced? What reagent is in excess and by how much? Determine which reagent will produce the smallest amount of product: 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 6.60 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑋 2 moles NH3 = 4.40 moles 𝑁𝐻3 3moles H2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 4.42 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 N2 𝑋 2 moles NH3 = 8.84 moles 𝑁𝐻3 1moles N2 Therefore the limiting reagent is 𝐻2.  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 The amount of limiting reagent present at the start of the reaction determines the theoretical yield. To determine the amount of NH3 produced, use the limiting reagent. 6.60 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑋 2 𝑚𝑜𝑙𝑒𝑠 NH3 = 4.40 moles NH3 (Theoretical yield) 3 moles H2 The excess reagent is N2. If you have 6.60 moles H2 then you will need 6.60 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑋 1 𝑚𝑜𝑙𝑒𝑠 N2 = 2.20 moles N2 3 moles H2  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 But you have 4.42 moles N2 Therefore, the excess amount of N2 is 4.42 moles – 2.20moles = 2.22 moles N2  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 b. If 25.5 g H2 are made to react with 64.2 g N2, what is the limiting reagent? What is the theoretical yield in g of NH3 that will be produced? How do you determine the limiting reagent? Get the number of moles of each reactant. Calculate the number of moles of product using each reagent. The one that yields the smallest number of moles of product is the limiting reagent.  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 3𝐻2(𝑔) + 𝑁2 (𝑔) → 2𝑁𝐻3 (𝑔) 25.5 g 𝐻2 𝑋 1 moles H2 = 12.6 moles 𝐻2 2.016 H2 64.2 g N2 𝑋 1 moles N2 = 2.29 moles N2 28.02 N2  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 3𝐻2(𝑔) + 𝑁2 (𝑔) → 2𝑁𝐻3 (𝑔) From 12.6 moles of H2, how many moles of NH3 do we expect to get? 12.6 moles 𝐻2 𝑋 2 moles NH3 = 8.40 moles N𝐻3 3 moles H2 From 2.29 moles of H2, how many moles of NH3 do we expect to get? 2.29 moles N2 𝑋 2 moles NH3 = 4.58 moles N𝐻3 1 mole N2 The limiting reagent is N2.  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 Note: If the amount of the initial reactant is expressed in grams (mass) instead of moles, the number of moles must first be converted into grams using the molar mass of the reactant. If the problem asks to determine the mass (in grams) of the product produced by the reaction, the number of moles of the product must be converted into grams using the molar mass of the product.  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 Let’s go online! Directions: Go to the link below and practice what you’ve learned from this lesson: https://bit.ly/309SojJ  Lesson 2: Calculating Theoretical Yield and Percent Yield in a Reaction  Encircle Me… Direction: In this chemical equation, identify and encircle the following parts. 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) Direction of the Reaction 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) Product/s 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) Reactant/s 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) Subscripts 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) Coefficients 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) Theoretical yield is the amount of product that is expected to form based on stoichiometry. It is the maximum amount of product produced from the given amount/s of reactant/s. It is calculated based on the stoichiometry of the chemical reaction. Actual yield is the amount of product produced during the reaction. It is the amount of product obtained after the actual reaction and it is normally lesser than the theoretical yield. It is determined experimentally. Yield of the Reaction Percent yield is the ratio of the actual yield to the theoretical yield expressed as a percentage. Following the formula of: 𝑷𝒆𝒓𝒄𝒆𝒏𝒕 𝒀𝒊𝒆𝒍𝒅 = 𝑨𝒄𝒕𝒖𝒂𝒍 𝒀𝒊𝒆𝒍𝒅 𝑿 𝟏𝟎𝟎 𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒀𝒊𝒆𝒍𝒅 If in the example given, only 54.0 g NH3 were produced, then the actual yield is 54.0 g; the theoretical yield is 78.0 g and the % yield is: % 𝒀𝒊𝒆𝒍𝒅 = 54.0 g = 69.2% 78.0 g Steps in calculating the percent yield of a chemical reaction: 1. Balance the given chemical equation 2. Identify the limiting reactant and the excess reactant 3. Compute for the theoretical yield of the reaction 4. Calculate the percent yield You can continue solving when asked: a. Percent error b. Amount of excess reactant  Activity 1: Solve, Solve, Solve… Dichlorodifluoromethane, CCl2F2, a refrigerant is prepared from reaction of CCl4 and HF. CCl4+2HF → CCl2 F2 +2HCl What happen when 3.00 grams of CCl4 reacts with 3.00 grams of HF? a. Which is the limiting reagent? b. How much of the excess reagent remains unreacted? c. How many grams of CCl2 F2 was produced? Procedure: Avogardo’s Avogardo’s Particles number mole mole ratio mole number Particles of A of A Coefficient of B of B X 1mol of A and B X 6.02X10 23 6.02X10 23 1 mol Use molar masses as Mass conversion Mass factor of A of B  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 Let’s go online! Direction: Refer to the following sites/links for further discussions on how to calculate the theoretical yield using stoichiometry and percent yield using the formula. https://bit.ly/3gEnN4f https://bit.ly/3edzFIM https://bit.ly/3faDK1W  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 Cite one (1) real-life situation where you added more than what is needed to produce something. Make justifications why you have to do it.  𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 Tell Me What You Know 1. What is the role of the stoichiometric coefficient in a balanced chemical equation to the determination of the limiting reactant? 2. Why is it that in most cases, the amounts of reactants and products are indicated in grams instead of the number for moles. 3. How do you determine the limiting reactant in a chemical reaction? 4. If a chemical reaction involves only one reactant, will there be a limiting reagent? Thank You!!!