Science, Tech, Math › Science Theoretical Yield Example Problem Calculate the Amount of Product Produced From Given Amount of Reactant Share Flipboard Email Print You can calculate the theoretical yield of a chemical reaction to know about how much product to expect. Ben Mills Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. our editorial process Todd Helmenstine Updated February 06, 2020 This example problem demonstrates how to predict the amount of product formed by a given amount of reactants. This predicted quantity is the theoretical yield. Theoretical yield is the amount of product a reaction would produce if the reactants reacted completely. Problem Given the reactionNa2S(aq) + 2 AgNO3(aq) → Ag2S(s) + 2 NaNO3(aq)How many grams of Ag2S will form when 3.94 g of AgNO3 and an excess of Na2S are reacted together? Solution The key to solving this type of problem is to find the mole ratio between the product and the reactant.Step 1 - Find the atomic weight of AgNO3 and Ag2S.From the periodic table:Atomic weight of Ag = 107.87 gAtomic weight of N = 14 gAtomic weight of O = 16 gAtomic weight of S = 32.01 gAtomic weight of AgNO3 = (107.87 g) + (14.01 g) + 3(16.00 g)Atomic weight of AgNO3 = 107.87 g + 14.01 g + 48.00 gAtomic weight of AgNO3 = 169.88 gAtomic weight of Ag2S = 2(107.87 g) + 32.01 gAtomic weight of Ag2S = 215.74 g + 32.01 gAtomic weight of Ag2S = 247.75 gStep 2 - Find mole ratio between product and reactantThe reaction formula gives the whole number of moles needed to complete and balance the reaction. For this reaction, two moles of AgNO3 is needed to produce one mole of Ag2S.The mole ratio then is 1 mol Ag2S/2 mol AgNO3 Step 3 Find amount of product produced.The excess of Na2S means all of the 3.94 g of AgNO3 will be used to complete the reaction.grams Ag2S = 3.94 g AgNO3 x 1 mol AgNO3/169.88 g AgNO3 x 1 mol Ag2S/2 mol AgNO3 x 247.75 g Ag2S/1 mol Ag2SNote the units cancel out, leaving only grams Ag2Sgrams Ag2S = 2.87 g Ag2S Answer 2.87 g of Ag2S will be produced from 3.94 g of AgNO3.