Download Limiting Reagent Tutorial

Tutorial on Limiting Reagents Concept
DEFINITION: The limiting reagent is the reactant that when used up prevents further products being formed
even though there is a surplus of other reactants in the mixture. Let’s look at a common back yard cookout
example: Most hotdogs come in packs of 10 and most hot dog buns come in packs of 8. As you know, to make
a complete sandwich you need to place one hotdog into one bun. So, how many hotdog sandwiches can you
make if you buy only one pack of hotdogs and one pack of buns? Yep, you can only make 8 sandwiches – and
you’ll have two hotdogs left over.
Chemical reactions are like this. As you know the reactants will come together in specific proportions. For
example, let’s look at the reaction of sodium and chlorine as shown in Equation 1.
EQ1: 2 Na + Cl2  2 NaCl
So for a complete reaction to take place, you would have to have exactly two sodium atoms and one molecule
of chlorine to make two molecules of sodium chloride. What if you had ten atoms of sodium and 5 chlorine
molecules? How many sodium chloride molecules would be made? Yes, you got it again: 10 molecules. So if
you looked just at the number of reacting units sodium would be the limiting factor.
But what if you had a 100 grams of each reactant? Which reactant would be the limiting factor? Since we have
seen from the above discussion that you need twice as many atoms of sodium as you do molecules of chlorine,
the question is how many atoms of sodium and molecules of chlorine are in 100 grams. We can reduce the
magnitude of the question by asking how many moles of each of the reactants are in 100 g. So Equation 2
calculates the number of moles of sodium in 100 g of sodium.
EQ2: n Na = mass Na/molar mass (AW) of Na = 100 g/22.99 g/n = 4.34972 n Na.
Equation 3 calculates the number of moles of chlorine gas in 100 g of the gas.
EQ3: n Cl2 = mass Cl2/molar mass (FW) Cl2 = 100 g /70.90 g/n = 1.41044 n Cl2
From the balanced equation for the reaction (EQ1) we find that 4.34972 n Na would react with exactly 2.1746 n
of Cl2 (Equation 4); since we know from EQ3 we have only 1.41044 n Cl2, it is obvious that chlorine is the
EQ4: 4.34972 n Na x n Cl2/ 2 n Na = 4.34972/2 n Cl2 = 2.1746 n Cl2.
limiting factor under these conditions. We can now calculate the mass of sodium left over when all the chlorine
has been used up using EQ 1 (Equation 5). Equation 6 shows how the number of moles of sodium left after the
reaction is over can be calculated.
EQ5: n Na consumed = 1.41044 n Cl2 x 2 n Na/ n Cl2 = 2.82088 n Na.
EQ6: n Na remaining = Total n Na – n Na Consumed = 4.34972 n Na (EQ2) - 2.82088 n Na (EQ5)
= 1.52884 n Na
Finally, Equation 7 shows how to calculate the mass of the sodium left using the inverse of EQ2 & EQ6.
EQ7: 1.52884 n Na x 22.99 g Na/n Na = 35.1480316 g Na. = 35.1 g (to proper significance)
Let’s do another example. Hydrogen burns in oxygen to yield water. Suppose you began the experiment with
5.00 n hydrogen and 2.30 n oxygen. We want to answer the following questions: 1) Which reactant is the
limiting reagent? 2) How many moles water was formed? 3) How many moles of the excess reactant are left.
A. First write and balance chemical equation for the reaction. Remember that both gases are diatomic.
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B. Find out how many moles of oxygen are consumed by 5.00 moles of hydrogen and decide which reactant is
limiting. (If that value is more than 2.30 n O2, then oxygen is the limiting factor; if that value is less than
2.30 n O2, then hydrogen is the limiting factor.)
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C. Now calculate the amount of moles of water produced using 2.30 n O2 (Since this will be the most water
that can be produced).
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D. Calculate how much hydrogen it it would take to produce the amount of water produced (answer to C) in the
reaction.
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E. Finally, calculate the amount of unused hydrogen.
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This concludes the tutorial.