limiting excess reactants with stoichiometry n.
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Limiting & Excess Reactants with Stoichiometry. Limiting Reactants. One of the two (or more) reactants will run out first. This LIMITS the amount of product produced, so ……………. Take each reactant and convert to product

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limiting reactants
Limiting Reactants
  • One of the two (or more) reactants will run out first. This LIMITS the amount of product produced, so …………….
slide3

Take each reactant and convert to product

  • Whichever reactant made the LEAST amt of product is the Limiting Reactant
example
Example:
  • First you are given a chemical equation.
  • C2H4 + O2 CO2 + H2O
  • Balance the Equation first
  • C2H4 + 3 O2 2 CO2+ 2H2O
you have 2 70 moles of c 2 h 4 and 6 30 moles of oxygen gas how many moles of water are formed
You have 2.70 moles of C2H4 and 6.30 moles of oxygen gas. How many moles of water are formed?
  • C2H4 + 3 O2 2 CO2 + 2H2O
  • Now draw boxes around the reactants and products that are involved with the data.
  • C2H4 + 3 O2 2 CO2 + 2H2O

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

slide6

C2H4 + 3 O2 2 CO2 + 2H2O

  • 2.70 mol 6.30 molx 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”
slide7

C2H4 + 3 O2 2 CO2 + 2H2O

  • 2.70 mol 6.30 mol x mol
  • 2.70 mol C2H42 mol H2O = 5.40 mol H2O
  • 1 1 mol C2H4
  • 6.30 mol O22 mol H2O = 4.20 mol H2O
  • 1 3 mol O2
slide8

One of the answers is correct and one in WRONG!

  • Correct Answer is the leastamt 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 H2O
  • What is the LR?
  • O2
how to solve for how much excess remains
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?)
excess continued
Excess continued…
  • Convert LR into excess reactant (this is how much is actually needed)
  • 6.30 mol O21 mol C2H4= 2.10 mol C2H4
  • 1 3 mol O2 (needed)
now to find out how much ex remains
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 C2H4 (given in the problem)

2.10 mol C2H4 (needed)

0.60 mol of EX reactant

summary table
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: ________
summary
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.
summary continued
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)