Welcome to Solution Stoichiometry!
Solution stoichiometry is concerned with reactions that are aqueous,
or, in other words, reactions that take place in water.
And this tutorial assumes you have an acquaintance with basic stoichiometry.
If you are not acquainted with basic stoichiometry you should first watch my basic stoichiometry video by clicking here before moving on to solution stoichiometry.
So aqueous reactions are occurring all around us and used in many ways.
Producing a metal from metal ore is an aqueous reaction. The bonding of cement to the stones mixed into concrete is an aqueous reaction.
The reactions occurring in the oceans, lakes, and streams on the earth are aqueous.
And we wouldn’t have any living organisms on the earth without aqueous reactions.
All the reactions occurring in living organisms are aqueous!
That’s one of the big reasons why you need so much water in your body!
Without enough water your reactions slow down and eventually stop! This would be an unfortunate situation to be in!
So water provides a medium in which these reactions can occur.
For example, adding solid cupper II chloride to solid aluminum results in no reaction.
But if water is added, it allows the copper ions to contact the aluminum, resulting in a reaction between the two.
An important vocabulary term is the solution itself.
It is composed of a mixture of solute and solvent; or the substance that gets dissolved mixed with the liquid that does the dissolving.
Reactions in solution are almost always in water, and so water is the solvent.
For example dissolving copper II choride in water would result in a solution of CuCl₂ , written as CuCl₂(aq).
The (aq) tells us it is a solution, it is aqueous.
What is critical to know for solution stoichiometry is concentration, or how much solute is dissolved per volume of solution.
There are various ways of measuring concentration, for solution stoichiometry concentration is normally expressed as molarity.
Molarity gives us the concentration as number of moles of solute per liter of solution, or mols per liter, symbolized as capital M, which is read as molar.
So if you see capital "M" representing molarity, it is read as "molar."
So an example would be a .30 molar solution of aluminum chloride,
which tells us there is a concentration of .30 mols of AlCl₃
dissolved for every one liter of solution.
Critical: Molarity gives us the numerical relationship of solute to solution, and so it gives us another way to get to moles!
Another conversion factor.
This is a good thing…. In stoichiometry you always need a way to get to moles.
Ok, now we can look at an example of solution stoichiometry:
When reacting sodium carbonate and aluminum chloride, what volume of 1.2 molar sodium carbonate is needed to react with all of 5 liters of .30 molar aluminum chloride?
Let’s inspect the information given in the problem. The coefficients are provided in the reaction and we cannot do stoichiometry without coefficients.
We are given two solutions in the problem, but these are unit relationships, both molarity, so those are not specific amounts of something, but are unit relationships that can be used to convert one unit to another.
The 5 liters of aqueous AlCl₃ is a specific amount of something, so we begin with that.
In stoichiometry always inspect the problem to separate unit relationships from some specific amount of something, which is the thing we start with.
We know in stoichiometry that we have to get to moles, and the molarity provides a way to do this.
Multiplying the 5 liters by the concentration of the aluminum chloride solution, 0.30 mols/liter, allows for the conversion to mols of aluminum chloride.
Note that the liters here represents the solution of AlCl₃ , liters of aqueous aluminum chloride, so they both cancel.
Now that we have mols of AlCl₃ we can use the mole-mole relationship given in the equation.
There are 3 sodium carbonates for every 2 aluminum chlorides. This coefficient ratio allows us to convert to moles of Na₂CO₃.
The problem is asking for a volume of sodium carbonate, so we can use the molarity of sodium carbonate to convert to volume: for every liter of aqueous Na₂CO₃, there is 1.2 mols of Na₂CO₃.
This allows us to cancel moles of Na₂CO₃ and we are left with volume of aqueous sodium carbonate. What is that telling us though? It’s telling us that it takes 1.875 liters of 1.2 molar Na₂CO₃ to react with all of 5.0 liters of .30 molar AlCl₃.
I hope you can see that these sorts of calculations are very useful in the lab to determine amounts of substances needed in a reaction.
Let’s take a look at the calculation set up: the first calculation gives us mols of AlCL₃, the second gives us mols of Na₂CO₃, and finally the last conversion gives us what is asked for, volume of aqueous Na₂CO₃.
See if you can do the next one:
Okay, let's see what we got here.
Here's a relationship, something that we can use for a conversion. It's not an amount.
Here's the amount given in the problem.
So you're starting with 35 grams of Aluminum.
How much FeCl₂ is going to react with that?
So you always want to get to moles somehow.
If you have grams of Al, you divide by the molar mass.
And it gives you moles of Al.
And then the coefficient ratio gives you moles of FeCl₂.
Now you're multiplying by the molarity.
For every one liter there is .45 moles of FeCl₂.
And that gives us a grand total of 4.32 L of aqueous FeCl₂.
So if you want to stop the video and make sure you have all of that written down, go ahead and do that.
And let's take ten minutes for completing problems 1, 2, and 3.
And these answers, I think, are correct.
And once you are done, turn the video back on and we will take a look at the work.
Okay. So here we have the balanced equation. And I hope you balanced it first.
Here is a relationship, a concentration. Another concentration is here.
So this is the amount we're given in the problem.
So we start off with 0.12 L of Fe(NO₃)₃
Convert that to moles of Fe(NO₃)₃. Then the mole ratio, and then the molarity of KF.
1 liter for every 2.2 moles of KF gives us a grand total of 0.139 L
Next one: The amount given is 5 L. Here is the molarity of the KF.
And we're looking for how much mass.
What's the mass, what's the weight of Fe(NO₃)₃ that we're going to have to add to this amount of KF?
So the volume of KF times the molarity gives us:
Here's our moles of KF.
The mole ratio from the equation gives us moles of Fe(NO₃)₃
And then you just multiply it by the molar mass,
giving us a grand total of 584.5 g
And then finally, something slightly more complicated.
We're looking at a relationship given by density.
Grams per cubic cm as well as molarity.
And we're given a volume of solid KF and asked how much Fe(NO₃)₃ will react with that.
So we have to get to moles from our volume, but we can't do that unless we get to grams.
So this calculation gives us grams of KF.
And then we can divide by the molar mass to get moles of KF.
The mole ratio and then the molarity gives us a grand total of 5.34 L
Alright. I hope you enjoyed this as much as I have.
I will see you very soon, I hope. Bye!