Lesson 28: Determining the Needed Amount of a Solution to Perform a Reaction
(See pages 460-466)
In Lesson 27 you learned how to take a known amount of solution of a known molarity and place it into a
chemical reaction in order to predict the quantity of a product from the reaction. You learned that you first have
to convert the milliliters of solution to grams of solute and then ultimately moles of solute before you could
continue with the stoichiometry calculations. In this lesson you will learn how to determine the amount of
known molarity solution necessary in order to produce a desired amount of product. This procedure is very
similar to determining the amount of reactant necessary to produce a desired amount of product (Lesson 25).
However, in this case instead of determining the amount of dry reactant needed, you will be asked how many
milliliters of a known molarity solution it will take to make the desired amount of product.
Let's begin by looking at an example:
Example 1: Suppose that you've been asked to prepare 100 grams of potassium chloride for your boss. You've
found in your store room that you have 5 M potassium chlorate which will readily decompose to produce potassium
chloride and oxygen gas. How many milliliters of 5 M potassium chlorate will be necessary to make the 100 grams of
potassium chloride?
Start by writing and balancing the chemical reaction.
KCl03
KCl + 02
Balance the equation and Identify the reaction components we are interested in.
? ml 5M
2KCl03
100 grams
2KCl + 302
Then, determine the mole ratio.
2:2 which is the same as 1:1
Next, let's look back at our problem. How many grams of potassium chloride are we supposed to make?
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Recall from Lesson 22 that we can only apply the mole ratio to moles (not grams) of product or reactant. So, the
next thing we'll need to do is convert these desired 100 grams of potassium chloride into moles of potassium chloride.
And, since you've done this many, many times, you know you'll need a formula weight for the potassium chloride.
KCl
K: 1 x 39.0 g = 39.0 grams
Cl: 1 x 35 g = 35.5 grams
Formula weight = 74.5 grams
Now, we've been asked to make 100 grams of potassium chloride from this reaction. Knowing one mole of
potassium chloride has a mass of 74.5 grams allows us to create this set of equivalent fractions:
1 mole KCl
x moles
74.5 grams = 100 grams
x = 1.35 moles KCl
or the equation method:
100 grams X 1 mole KCl
74.5 grams
=
1.35 moles KCl
NOTICE WE ARE ALWAYS CONVERTING BACK AND FORTH BETWEEN MOLES AND GRAMS. THIS IS BECAUSE
THE MOLE RATIO ONLY WORKS WITH MOLES. SO WHEN YOU START A PROBLEM THE FIRST THING YOU DO IS
FIND MOLES. NOTICE THE MOLE-MOLE PROBLEMS ARE THE EASIEST. YOU ALREADY HAVE MOLES SO YOU CAN
APPLY THE MOLE RATIO IMMEDIATELY - THEN YOUR ANSWER ASKS FOR MOLES AND SINCE YOU ALREADY
HAVE MOLES YOU ARE DONE. NO CONVERSIONS REQUIRED!
Therefore based upon these calculations, we see we'll need to produce 1.35 moles of potassium chloride (KC1).
Since we have this amount now in moles, we can apply the mole ratio we found earlier.
1:1
So, based upon this ratio, we know in order to produce 1.35 moles of potassium chloride, it will require 1.35
moles of potassium chlorate.
Now, check back to our problem. We are being asked to determine how many milliliters of a 5 M potassium
chlorate solution are necessary to make the desired 100 grams of potassium chloride. If our available solution is 5 M,
that means that there are 5 moles of potassium chlorate per liter (1000 ml) of solution. Knowing that, we can set up
these two equivalent fractions:
In 1000 milliliters solution
There are 5 moles of KClO3
How many (x) milliliters
in 1.35 moles of KClO3
=
5 x = 1350
x = 270 ml KClO3
OR
1.35 moles KClO3
X
1000 milliliters solution KClO3
5 moles KClO3
=
270 ml KClO3
Therefore, based upon these calculations, we can say that in order to prepare 100 grams of
potassium chloride, we would need at least 270 ml of 5 M potassium chlorate.
Let's look at another example:
Example 2: If bromine gas is bubbled through a solution of potassium iodide, potassium bromide and
iodine crystals are produced. Suppose that you've been asked to prepare 50.0 grams of iodine crystals (I2)
from this reaction. If the potassium iodide solution is 2.0 M, how many milliliters of this solution would you
need?
Begin by writing and balancing the chemical reaction:
Br 2 + 2KI
2KBr + I 2
Locate the parts of the reaction that you are concerned with:
? ml 2.0 M
50 grams
Br2 + 2KI
2KBr +
I2
And determine the mole ratio:
2:1
Look back at the problem now. We are being asked to prepare 50.0 grams of iodine
crystals. Before we can apply the mole ratio to this desired amount of product, we must first
convert it into moles. To do this we need to determine how many grams are in 1 mole of iodine
crystals aka finding the formula weight of I2.
I:
2 x 127 g = 254 grams / mole
Knowing this amount, we can then set up equivalent fractions to find out how many moles
are equal to the 50.0 grams we are being asked to prepare.
1 mole I2
254 g
=
x moles
50.0 g
x = 0.200 moles Iodine
OR
50.0 grams Iodine X
x moles iodine
254 grams Iodine
=
0.200 moles Iodine
Based upon the calculations, we see that our 50 grams of iodine crystals are equivalent to 0.20 moles of iodine
crystals. Now that we have this amount in moles, we can apply it to the mole ratio:
2:1
0.400 moles KI —> 0.200 moles I2
In order to make the desired 0.20 moles of iodine crystals, it will take twice as many moles of potassium iodide
or 0.40 moles.
Recall that our potassium iodide is found in a 2 M solution. We know we will need 0.400 moles, but now
need to know how many milliliters that is equivalent to. A 2M solution means that there are 2 moles of KI found
in each liter (1000 ml). Knowing this we can set up two equivalent fractions:
In 1000 ml there are
2 moles KI
=
How many ml (x)
0.40 mole KI
2x = 400
x = 200 ml KI
OR
0.40 mole KI
X
1000 ml KI
2 moles KI
=
200 ml KI
Based upon these calculations, we can say that it will take 200 ml of 2 M KI in order to make the desired amount
of 50 grams of iodine crystals.
To review what you've learned in these examples, first we attacked the problem just like any other
stoichiometry problem. We wrote and balanced the chemical reaction. We determined the mole ratio and used it
to predict the amount of ingredient necessary to produce the desired amount of product. We then looked at the
molarity of the reactant solution we had available and finally determined the volume necessary to meet our
needs in the reaction.
Determining Volume of Reactant Necessary to Produce Desired Product
1. Joe has been asked to prepare 300 grams of hydrogen cyanide for his boss. He is going to use the following reaction:
Hydrogen sulfate and sodium cyanide react to yield hydrogen cyanide and sodium sulfate. He has 3 M hydrogen
sulfate on hand for the reaction. How many milliliters of this solution will be required to make the 300 grams of HCN?
2. Patricia was preparing to make 450 grams of zinc chloride. She knew that when zinc oxide and hydrochloric acid
are combined, they product zinc chloride and water. She looked on her supply shelf and found she had 5 M
hydrochloric acid and plenty of zinc oxide. How many milliliters of the 5 M hydrochloric acid will she need?
3. Terry knew that carbon and aluminum oxide will react to produce aluminum and carbon dioxide. If Terry had a
container of 5 M aluminum oxide in his supply cupboard, how many milliliters would it take for him to prepare 500
grams of aluminum based upon his reaction?