What is It Unlocking of Difficulties Chemical Formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule, using chemical element symbols, numbers, and sometimes also other symbols, such as parentheses, dashes, brackets, commas and plus (+) and minus (-) signs. Chemical equation is the symbolic representation of a chemical reaction in the form of symbols and formulae, wherein the reactant entities are given on the left-hand side and the product entities on the right-hand side Molar mass is a physical property, defined as the mass of a given element or molecule per mole of that substance. Since a mole is defined as the amount of a substance and substances have different masses, each element or molecule will have a different molar mass Molecular weight is a measure of the sum of the atomic weight values of the atoms in a molecule. Limiting reactant is a reactant that is completely used up in a chemical reaction. Excess reactant is a reactant that still remains after the reaction stops. Actual yield is the amount of product actually obtained in a chemical reaction. Theoretical yield is the amount of product that could possibly be produced in a chemical reaction. It can be calculated through stoichiometry. Percent yield refers to the efficiency of a chemical reaction. It is the ratio of the actual yield to the theoretical yield multiplied by 100. The formula for calculating the percent yield is actual yield Percent yeild = theoretical yield X 100* Determining the Limiting and Excess Reactant A chemical reaction stops when one of the reactants is completely used up. In most chemical experiments, the reactants are not in exact ratios specified by the balanced chemical equation. Often, one of the reactants is used in excess, and the reaction is allowed to proceed until one of the reactants is used up. To determine the limiting reactant, calculate the number of moles of product that will be produced from the given reactants. The reactant that will yield a smaller amount of product is the limiting reactant. Follow Me! Sample Problem no. 1 Consider the reaction between potassium oxide and hydrochloric acid. K20 + 2 HCI - 2 KCI + H20 If 10.0g K20 was made to react with 10.0g of HCl, which reactant is limiting? Which reactant is in excess? (Use the following molar masses: K20 = 94.2 g/mol; HCI = 36.5 g/mol.) Solution: Step 1: Convert the mass of both reactants to their corresponding moles. mass K20 moles K20 = - molar mass K20 10.0 g moles K20 = 94.2g/mol mass K20 moles K20 = molar mass K20 moles K20 = 0. 1062 mol of K20 are present in the reactant. mass HCI moles HCI = molar mass HCI 10.0 g moles HCI = 36.5g/mol moles HCI = 0.2739 mol of HCl are present in the reactant.Step 2: Calculate the number of moles of product that can be formed from each reactant. You may use either H20 or KCI. Let x = amount of H20 formed from 0.106 mol of K20 1 mol K20 0.106 mol K20 1 mol H,0 X Rearrange the equation, and solve for x. (0.106 mol K20) (1 mol H20) X = (1 mol K20) x = 0.106 mol H20 would be produced. Let y = amount of H20 formed from 0.274 mol of HCI 2 mol HCI 0.274 mol HCI 1 mol H20 y Rearrange the equation, and solve for y. y = (0.274 mol HCI) (1 mol H20) (2 mol HCI) y = 0.137 mol H20 would be produced. Step 3: Compare the number of moles of product formed from each reactant. . The number of moles of H20 produced from K20 is less than the number of moles of H20 produced from HCI. Step 4: Tag the reactant that gives a lower amount of product as the limiting reactant. o K20 is the limiting reactant. Step 5: Tag the reactant that gives a greater amount of product as the excess reactant. o HCl is the excess reactant.* Calculating the Theorical Yield and Percent Yield. In calculating the percent yield, you need to calculate first the theoretical yield and then compare it with the actual yield. The answer is then expressed in percentage. A higher percentage indicates a more efficient chemical reaction. Follow Me! Sample Problem no.2 Sodium hydroxide reacts with sulfuric acid to produce sodium sulfate and water. 2 NaOH + H2SO4 - NazSO4 + 2 H20 . What is the theoretical yield (in grams) of NazSO, that will be formed when 4.00 mol of NaOH are completely consumed in the reaction? (NazSO4 has a molar mass of 142g/mol.) . What is the percent yield if only 270g were produced? Solution: Step 1: Balance the chemical equation. 2 NaOH + H2SO4 - Na2SO4 +2 H20 Step 2: Determine the mole ratio between NaOH and Na,SO4. 2mol NaOH 1molNa2 SO4 Step 3: Set up an equation to solve for the unknown. Let x = amount of NazSO4 formed from 4.00 mol of NaOH 2 mol NaOH 4.00 mol NaOH 1mol Na2 SO4 x Step 4: Rearrange the equation, and solve for x. (4.00 mol NaOH)(1mol Na2504) * = (2 mol NaOH) x = 2.00 mol Na2504Step 5: Convert the given moles to mass. moles Na2SO4 = mass Na2SO4 molar mass Na2SO4 mass Na2504 = moles Na2504 x molar mass Na2504 mass Na2504 = 2.00 mol x 142 g/mol mass Na2504 = 284 g Therefore, the reaction produces 284 gof NazSO4 when 4.00 mol of NaOH was completely consumed. The theoretical yield is 284 g NazSO4 a. Step 1: Identify the actual and theoretical yields. Actual yield = 270g NazSO4 Theoretical yield = 284g Na2SO. Step 2: Use the formula to calculate percent yield. actual yield percent yield = = theoretical yield x 100 percent yield = 270g x 100 284g percent yield = 95. 1% Therefore, the reaction is 95.1% efficient. In a Nutshell! Calculating the Percent Yield of the Reaction Based on the Amount of Limiting Reactant Write and Convert the mass of Calculate the number both reactants to balance the of moles of product their corresponding that can be formed equation. moles. from each reactant. Convert the moles of Tag the reactant that Compare the number product formed from gives a lower amount of of moles of product the limiting reactant to product as the limiting formed from each mass. reactant reactant. Calculate the percent yield of the reaction using Identify the given equation: actual yield. actual yield percent yield = - theoretical yield x 100 1312 De DEPARTMENT OF EDUCATION Physical Science Quarter 1 - Module 10: Limiting Reactant and Products Formed in a Reaction Module 1 Module 2 Module ] SELF- LEARNING MODULE 1 GOVERNMENT PROPERTY DEPARTMENT OF EDUCATION - SOCCERSARGEN NOT FOR SALELesson Limiting Reactant and Products Formed in a 6 Reaction This section describes how a reactant may limit' a chemical reaction, meaning, how one reactant may determine how much of the other substance used in the reaction and how much of the product can be formed. It also discusses why the actual yield of the product of a reaction may be less than expected. What's In Mabuhay! How are you today! Take a look around you. Did you ever wonder how plants produce their own food? How cheese, wine, beer, yogurt and bread and many other products you see everyday are produced? Did you ever try to strike a match, burn a candle, build a fire or light a grill? What do you notice? Matter interacts to form new products through process called chemical reaction or chemical change. Every time you clean or cook, its chemistry in action. Chemical change is constantly taking place in the environment every day. The following words may refer to chemical change: rust, burn, explode, decompose, corrode, ferment and cooking In cooking, we follow a set of recipe and directions to obtain the product that we want. In chemistry, a chemical equation is composed of the amount of materials used [reactants] and the new desired substance formed [product].7. Consider the reaction between 0.1mole Al203 and 1 mole H2. Al203+3H2 - 2Al + 3H20 Question: Which of the following is the limiting reactant? a.Al203 b. H2 c. H20 d. Al 8. Consider the reaction below. 3N2+ 302 -> 2N20 + N204 Question: If 3 Moles of N2 is used, how many moles Oz should be present so that Oz becomes the limiting reactant? a. 2 b. 4 c. 6 d. 8 9. Consider the reaction below. 3N2 + 302 -> 2N20 + N204 Question: Which of the following is the limiting reactant when 4 moles N2 reacts with 3moles O2? a. N2 b.N204 c. N20 d. O2 10. Consider the reaction between acetic acid and 10.6 grams' soda ash. Their action produced 13.1 grams NaCH,COO. 2CH3COOH + Na2CO3 + H20 + CO2 + 2NaCH3COO Use the following molar mass: Na2CO3 = 106 g/mol and NaCH3COO = 82 g/mol. Question: What is the percent yield of the reaction? a.63% b. 80% c. 100% d. 124%