--- title: "Reacting Masses, Limiting Reactants and Yield" description: "Use the mole ratio in a balanced equation to work out what mass of product a given mass of reactant can make, find which reactant runs out first, and compare the mass actually collected with the maxim" canonical: https://lightmysky.com/learn/science/reacting-masses-limiting-reactants-and-yield-mt_C2K6kp5Chv source: https://lightmysky.com/learn/science/reacting-masses-limiting-reactants-and-yield-mt_C2K6kp5Chv.md retrieved: 2026-09-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Reacting Masses, Limiting Reactants and Yield Use the mole ratio in a balanced equation to work out what mass of product a given mass of reactant can make, find which reactant runs out first, and compare the mass actually collected with the maximum. Subject: Science · Area: Matter & Materials · Ages 15 to 16 Page: https://lightmysky.com/learn/science/reacting-masses-limiting-reactants-and-yield-mt_C2K6kp5Chv ## Ready when they can - Runs a mass to moles to ratio to mass calculation for a named reaction - Identifies the limiting reactant from the moles available and the equation - Calculates percentage yield from the mass collected and the theoretical mass - Gives reasons a real yield falls short, such as losses on transfer, an incomplete reaction, or a side reaction ## Lesson: Reacting masses, limiting reactants and yield The equation the club balanced at the bench last week is a recipe written in particles. 2Mg + O₂ → 2MgO says two magnesium atoms for every one oxygen molecule, and two moles for every one mole. What it does not say is grams, because a mole of magnesium and a mole of oxygen weigh different amounts. So a balance reading cannot be fed into the ratio directly. Every calculation in this stop is the same detour: turn grams into moles, use the ratio, and turn moles back into grams. The route has four steps and they always run in the same order. First, mass to moles: divide the mass you were given by that substance's Mr. Second, read the ratio off the balanced equation. Third, apply it to get the moles of the substance you want. Fourth, moles to mass: multiply by the Mr of that substance. Skip the first step and you are doing arithmetic on grams, which the equation never promised anything about. *(drawing: Grams in, grams out, but the middle of the journey is always in moles.)* **Example.** Burn 2.4 g of magnesium ribbon. The equation is 2Mg + O₂ → 2MgO, with the relative atomic mass (Ar) of magnesium 24 and the Mr of magnesium oxide 40. Step one: 2.4 / 24 = 0.1 moles of magnesium. Step two: the ratio of Mg to MgO is 2 to 2, which is 1 to 1. Step three: 0.1 moles of magnesium gives 0.1 moles of magnesium oxide. Step four: 0.1 x 40 = 4.0 g. So 2.4 g of ribbon can make at most 4.0 g of ash. Now suppose both reactants are present and one runs out first. That one is the limiting reactant and it decides how much product you get, while the other is in excess and some of it is left over. Divide each reactant's moles by its number in the equation, and the smallest answer is the one that runs out. Then carry on from that one alone. For 2H₂ + O₂ → 2H₂O with 0.4 moles of hydrogen and 0.3 of oxygen: 0.4 / 2 = 0.2 against 0.3 / 1 = 0.3, so hydrogen limits, and 0.2 x 2 = 0.4 moles of water. *(drawing: Using all 0.3 moles of oxygen would take 0.6 moles of hydrogen, and only 0.4 is in the flask, so the hydrogen runs out first.)* **Example.** The 4.0 g from the ribbon is the theoretical yield, the most the reaction could give. Weigh the ash and you might collect 3.6 g. Percentage yield compares the two: divide the mass collected by the theoretical mass, then multiply by 100. Here 3.6 / 4.0 = 0.9, and 0.9 x 100 = 90 per cent. It falls short for ordinary reasons. Some product sticks to the container or is lost on transfer, the reaction may not run to completion, and some reactants may go off and make something else instead. **Recap.** An equation's numbers are a ratio of moles, never of grams, so every calculation runs mass to moles, ratio, moles to mass. When both reactants are present, divide each one's moles by its number in the equation; the smallest answer runs out first and caps the product. Percentage yield is the mass collected divided by the theoretical mass, times 100, and it falls short because product is lost, reactions do not finish, and side reactions happen. *Adapted from OpenStax Chemistry 2e (CC-BY 4.0), openstax.org* · [license](https://creativecommons.org/licenses/by/4.0/) ## Practice 28 questions on this page, each with its working shown. ## Needs first - [Relative Formula Mass and the Mole](https://lightmysky.com/learn/science/relative-formula-mass-and-the-mole-mt_-SiMKJK4rH) - [Proportion](https://lightmysky.com/learn/mathematics/proportion-mt_5mIcmKRCgA) - [Balancing Equations and Conservation of Mass](https://lightmysky.com/learn/science/balancing-equations-and-conservation-of-mass-mt_KUaPCEIA5f) ## Opens up - [Empirical and Molecular Formulas from Composition](https://lightmysky.com/learn/science/empirical-and-molecular-formulas-from-composition-mt_4rN5MQaTf7) - [Gas Volumes and the Ideal Gas Equation](https://lightmysky.com/learn/science/gas-volumes-and-the-ideal-gas-equation-mt_I1fz_6wjuD) - [Concentration of Solutions in Moles per Cubic Decimetre](https://lightmysky.com/learn/science/concentration-of-solutions-in-moles-per-cubic-decimetre-mt_vRK78ynnXb)