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Limiting & Excess Reactants with Stoichiometry

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Limiting Reactants One of the two (or more) reactants will run out first. This LIMITS the amount of product produced, so …………….

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Take each reactant and convert to product Whichever reactant made the LEAST amt of product is the Limiting Reactant

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Example: First you are given a chemical equation. C 2 H 4 + O 2 CO 2 + H 2 O Balance the Equation first C 2 H O 2 2 CO H 2 O

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You have 2.70 moles of C 2 H 4 and 6.30 moles of oxygen gas. How many moles of water are formed? C 2 H O 2 2 CO H 2 O Now draw boxes around the reactants and products that are involved with the data. C 2 H O 2 2 CO H 2 O Then write the givens and unknowns in. If the given is any unit BUT a mole, write it on top of the box (“house”). Moles are written on the bottom

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C 2 H O 2 2 CO H 2 O 2.70 mol 6.30 mol x mol Then draw arrows from where you are starting and where you are going. Now you are ready to do your multistep conversions! Convert each reactant to the unknown “x”

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C 2 H O 2 2 CO H 2 O 2.70 mol 6.30 mol x mol 2.70 mol C 2 H 4 2 mol H 2 O = 5.40 mol H 2 O 1 1 mol C 2 H mol O 2 2 mol H 2 O = 4.20 mol H 2 O 1 3 mol O 2

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One of the answers is correct and one in WRONG! Correct Answer is the least amt of product made! Whichever reactant made the correct amt of product is the LR. Going back to our problem, what is the correct answer for the number of moles of water produced? 4.20 mol of H 2 O What is the LR? O 2

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How to solve for how much Excess remains… To solve you need to see how much excess reactant is actually needed. (Think if you are making chocolate chip cookies, you look through what you have and you see that you have 1 bag of chips, a dozen eggs, and the right amt of all other ingredients needed. What do you have in excess? How much of the excess is actually needed for the recipe?)

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Excess continued… Convert LR into excess reactant (this is how much is actually needed) 6.30 mol O 2 1 mol C 2 H 4 = 2.10 mol C 2 H mol O 2 (needed)

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Now to find out how much EX remains….. Subtract what you just calculated (needed) from the given amt of excess (given in the problem) _ 2.70 mol C 2 H 4 (given in the problem) 2.10 mol C 2 H 4 (needed) 0.60 mol of EX reactant

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Summary Table At the end of each problem you want to have a summary of all of the work you have done. LR: ________ Product:________ EX Remains: ________

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Summary Show on the equation the “House Method” with arrows marking your way. Convert each reactant to the “x” (product or another reactant) One answer will be right- one will be wrong The correct answer is the one that has the least “x” made. The reactant that made that amt. of product is the LR.

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Summary Continued To solve for the excess, convert the LR into the EX reactant (temporarily disregard the amt of the EX in the given). This gives you the correct amt of EX that is needed for the reaction. Then Subtract what you just calculated from the given amt. of Ex. (Big – Small)

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