Limiting Reagent

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Even before beginning, a slight digression

You will see the word "excess" used in this section and in the problems. It is used several different ways:

a) "compound A reacts with an excess of compound B" - In this case, mentally set compound B aside for the moment. Since it is "in excess," this means there is more than enough of it. Some other compound (maybe A) will run out first.

b) "20 grams of A and 20 grams of B react. Which is in excess?" What we will do below is find out which substance runs out first (called the limiting reagent). Obviously (I hope), the other compound is seen to be in excess.

c) "after 20 gm. of A and 20 gm. of B react, how much of the excess compound remains?" To answer this problem, we would subtract the limiting reagent amount from the excess amount.

What is the Limiting Reagent?

It is simply the substance in a chemical reaction that runs out first. It seems to simple, but it does cause people problems. Let's try a simple example.

Reactant A is a test tube. I have 20 of them.
Reactant B is a stopper. I have 30 of them.

Product C is a stoppered test tube.

The reaction is: A + B ---> C
or: test tube plus stopper gives stoppered test tube.

So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up).

Suddently, we run out of one of the "reactants." Which one? That's right. We run out of test tubes first. Seems obvious, doesn't it? We had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted.

So, which "reactant" is limiting and which is in excess?

An Exercise in Limiting Reagent Here is what you know

  1. A cup of coffee costs 50 cents.
  2. In your possession, you have 100 grams each:
  3. You know how much a single coin weighs:

How many cups of coffee can you buy? The point to this type of problem is that it models what you have to do in chemistry. How do you figure out HOW MANY coins there are? That's right. You divide the total weight by the weight per one coin. You counted how many by weighing.

Chemistry does the same thing. I divide my total mass by the weight of one unit. In the example it is a coin, in chemistry it is the mole.

Limiting Reagent Problems

Problem #1: Here's a nice limiting reagent problem we will use for discussion. Consider the reaction:

2 Al + 3 I2 ------> 2 AlI3

Determine the limiting reagent and the theoretical yield of the product if one starts with:

a) 1.20 mol Al and 2.40 mol iodine.
b) 1.20 g Al and 2.40 g iodine
c) How many grams of Al are left over in part b?

Solution for part a: we already have moles as the unit, so we use those numbers directly.

Here is how to find out the limiting reagent: take the moles of each substance and divide it by the coefficient of the balanced equation. The substance that has the smallest answer is the limiting reagent.

Let's say that again:

to find the limiting reagent, take the moles of each substance and divide it by the coefficient of the balanced equation. The substance that has the smallest answer is the limiting reagent.

You're going to need that technique, so remember it.

For aluminum: 1.20 / 2 = 0.60
For iodine: 2.40 / 3 = 0.80

The lowest number indicates the limiting reagent. Aluminum will run out first in part a.

Why? 1.20/2 means there are 0.60 "groupings" of 2 and 2.40/3 means there are 0.80 "groupings" of 3. If they ran out at the same time, we'd need one "grouping" of each. Since there is less of the "grouping of 2," it will run out first.

If you're not sure what I just said, that's OK. The technique works, so remember it and use it.

The second part of the question "theoretical yield" depends on finding out the limiting reagent. Once we do that, it becomes a stoichiometric calculation.

Al and AlI3 stand in a one-to-one molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI3. Notice that the amount of I2 does not play a role, since it is in excess.

Solution for part b: since we have grams, we must first convert to moles. The we solve just as we did in part a just above.

For the mole calculation:

aluminum is 1.20 g / 26.98 g mol¯1 = 0.04477 mol
iodine is 2.4 g / 253.8 g mol¯1 = 0.009456 mol

To determine the limiting reagent:

aluminum is 0.04477 / 2 = 0.02238
iodine is 0.009456 / 3 = 0.003152

The lower number is iodine, so we have identified the limiting reagent.

Finally, we have to do a calculation and it will involve the iodine, NOT the aluminum.

I2 and AlI3 stand in a three-to-two molar relationship, so 0.009456 mol of I2 produces 0.006304 mol of AlI3. Again, notice that the amount of Al does not play a role, since it is in excess.

From here figure out the grams of AlI3 and you have your answer.

Solution for part c: since we have mole, we calculate directly and then convert to grams.

Al and I2 stand in a two-to-three molar relationship, so 0.009456 mol of I2 uses 0.006304 mol of Al.

Convert this aluminum amount to grams and subtract it from 1.20 g and that's the answer.


Problem #2: 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is the balanced equation for the reaction:

Al2S3 + 6 H2O ---> 2Al(OH)3 + 3 H2S

(A) Which is the limiting reagent?
(B) What is the maximum mass of H2S which can be formed from these reagents?
(C) How much excess reagent remains after the reaction is complete?

The key to this problem is the limiting reagent, part A. Once you know that, part B becomes "How much H2S can be made from the limiting reagent?" Part C becomes two connected questions: first, "How much Al2S3 is used up when reacting with the limiting reagent?" then second, "What is 15.00 minus the amount in the first part?"

Make sure you note that second part. The calculation gives you the answer to "How much reacted?" but the question is "How much remained?" Lots of students forget to do the second part (the 15 minus part) and so get graded down.

Here's the solution.


Problem #3: If there is 35.0 grams of C6H10 and 45.0 grams of O2, how many grams of the excess reagent will remain after the reaction ceases?

6 C6H10 + 17 O2 ---> 12 CO2 + 10 H2O

Solution:

1) Convert each substance to moles:

C6H10: 35.0 g / 82.145 g/mol = 0.426 mol
O2: 45.0 g / 31.998 g/mol = 1.406 mol

2) Determine the limiting reagent:

C6H10: 0.426 mol / 6 = 0.071
O2: 1.406 mol / 17 = 0.083

C6H10 is the limiting reagent.

3) Determine how many moles of the excess reagent is used up when the limiting reagent is fully consumed:

the mole ratio we desire is 6/17 (C6H10 to O2)

6/17 equals 0.426/x

x = 1.207 mol of O2 consumed

4) Determine grams of O2 remaining:

1.406 mol minus 1.207 mol = 0.199 mol of O2 remaining

0.199 mol times 31.998 g/mol = 6.37 g remaining


Problem #4: Based on the balanced equation:

C4H8 + 6O2 ---> 4CO2 + 4H2O

calculate the number of excess reagent units remaining when 28 C4H8 molecules and 228 O2 molecules react?

Solution:

Remember, numbers are just like moles, so treating the 28 and 228 as moles is perfectly acceptable.

1) Determine the limiting reagent:

butane: 28/1 = 28
oxygen: 228 / 6 = 38

Butane is the limiting reagent.

2) Determine how much oxygen reacts with 28 C4H8 molecules:

the butane:oxygen molar ratio is 1:6

28 x 6 = 168 oxygen molecules react

3) Determine excess oxygen:

228 - 168 = 60

Here's aother way to consider this:

The 38 above means that there are 38 "groupings" of six oxygen molecules.

38 minus 28 = 10 oxygen "groupings" remain after the butane is used up

10 x 6 = 60


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