Science, Tech, Math › Science How to Calculate Limiting Reactant and Theoretical Yield Share Flipboard Email Print Know the limiting reactant to calculate theoretical yield. Arne Pastoor/Getty Images 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 Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. our editorial process Facebook Facebook Twitter Twitter Anne Marie Helmenstine, Ph.D. Updated February 05, 2018 The limiting reactant of a reaction is the reactant that would run out first if all the reactants were to be reacted together. Once the limiting reactant is completely consumed, the reaction would cease to progress. The theoretic yield of a reaction is the amount of products produced when the limiting reactant runs out. This worked example chemistry problem shows how to determine the limiting reactant and calculate the theoretical yield of a chemical reaction. Limiting Reactant and Theoretical Yield Problem You are given the following reaction: 2 H2(g) + O2(g) → 2 H2O(l) Calculate: a. the stoichiometric ratio of moles H2 to moles O2b. the actual moles H2 to moles O2 when 1.50 mol H2 is mixed with 1.00 mol O2c. the limiting reactant (H2 or O2) for the mixture in part (b)d. the theoretical yield, in moles, of H2O for the mixture in part (b) Solution a. The stoichiometric ratio is given by using the coefficients of the balanced equation. The coefficients are the numbers listed before each formula. This equation is already balanced, so refer to the tutorial on balancing equations if you need further help: 2 mol H2 / mol O2 b. The actual ratio refers to the number of moles actually provided for the reaction. This may or may not be the same as the stoichiometric ratio. In this case, it is different: 1.50 mol H2 / 1.00 mol O2 = 1.50 mol H2 / mol O2 c. Note that the actual ratio of smaller than the required or stoichiometric ratio, which means there is insufficient H2 to react with all of the O2 that has been provided. The 'insufficient' component (H2) is the limiting reactant. Another way to put it is to say that O2 is in excess. When the reaction has proceeded to completion, all of the H2 will have been consumed, leaving some O2 and the product, H2O. d. Theoretical yield is based on the calculation using the amount of limiting reactant, 1.50 mol H2. Given that 2 mol H2 forms 2 mol H2O, we get: theoretical yield H2O = 1.50 mol H2 x 2 mol H2O / 2 mol H2 theoretical yield H2O = 1.50 mol H2O Note that the only requirement for performing this calculation is knowing the amount of the limiting reactant and the ratio of the amount of limiting reactant to the amount of product. Answers a. 2 mol H2 / mol O2b. 1.50 mol H2 / mol O2c. H2d. 1.50 mol H2O Tips for Working This Type of Problem The most important point to remember is that you are dealing with the molar ratio between the reactants and products. If you are given a value in grams, you need to convert it to moles. If you're asked to supply a number in grams, you convert back from the moles used in the calculation.The limiting reactant isn't automatically the one with the smallest number of moles. For example, say you have 1.0 moles of hydrogen and 0.9 moles of oxygen in the reaction to make water. If you didn't look at the stoichiometric ratio between the reactants, you might choose oxygen as the limiting reactant, yet hydrogen and oxygen react in a 2:1 ratio, so you'd actually expend the hydrogen much sooner than you'd use up the oxygen.When you're asked to give quantities, watch the number of significant figures. They always matter in chemistry!