Solution Stoichiometry: Videos & Practice Problems

Stoichiometry is a crucial concept in chemistry that involves using a balanced chemical equation to determine the relationship between the quantities of reactants and products. When incorporating solutions into stoichiometry, volume and molarity become important factors. For instance, if given the volume and molarity of water in a reaction with sodium to produce hydrogen gas and sodium hydroxide, one can calculate the grams of hydrogen gas produced. The process involves converting volume to liters if necessary, using molarity to find the moles of the given substance, and then using the mole ratio from the balanced equation to find the moles of the desired product. Finally, the moles of the product can be converted into grams using the molar mass. This approach is an extension of stoichiometry that now accounts for the volume and molarity of solutions in the calculations.

Solution Stoichiometry

Solutions stoichiometry deals with stoichiometric calculations in solutions that involve volume and molarity. So if you've watched my videos on stoichiometry, you are pretty familiar with our stoichiometric chart. Now that we are including solutions stoichiometry, we're going to adapt it slightly to fit the situations where now volume and molarity are sometimes given.

So here in our solutions stoichiometric chart, we're going to say a chart uses the given quantity of a compound to determine the unknown quantity of another compound. So for here in this example, we have our balanced chemical equation and it says for every two moles of sodium solid reacting with two moles of water as a liquid, we produce one mole of hydrogen gas and two moles of sodium hydroxide as an aqueous product. Now they're giving to us that water has a volume of 38.74 milliliters of 0.275 molar. So this is our given information and then what they're asking us to do is to determine the grams of H₂.

Now remember in our stoichiometric chart on the left side is where we have our given information. We are used to seeing our given information given to us in grams or in moles. But now it's solutions stoichiometry, we'll tend to see it also be given to us as volume of some molarity. And remember we've said this before. The word of in between 2 numbers means multiply and realize here that moles equals liters times molarity. So if you were to change these mL to liters and multiply them by the molarity then you would see that this new given shape that we are seeing feeds directly into our moles of given.

And remember once we have our moles of given it's our job to get to our unknown. Our unknown is what we're looking to find here. Our unknown is H₂ and we need to find it in grams. So following the stoichiometric chart we have here, we go from the molar given which is H₂O and we do the jump. In the jump, we're going from an area where we have information we know, to an area where we don't know anything about the unknown. And in this jump we have to remember to do a mole to mole comparison where we use the coefficients in the balanced equation.

At this point, it would take us to moles of H₂ and then it's up to us to go to either grams of H₂ or ions, atoms, formula units or molecules of whatever our unknown would be. So again, this is a continuation of our idea with stoichiometry, but now we're including this new shape here. If you haven't watched my video on stoichiometry, I highly suggest you go back and take a look at a couple of those videos. It's just a continuation of that idea now including molarity and volume within our calculations.

Now that we've seen this new stoichiometric chart, let's move on to some questions where we put it to practice.

Solution Stoichiometry Example

In this example question, it says how many moles of hydrogen gas were produced when 38.74 milliliters of 0.275 molar water reacts with excess sodium. All right, so we know that this is a balanced chemical equation. They're giving us information on one compound and asking for information on another. So this is the definition of stoichiometry balancing chemical equation. They give us information on at least one compound and we use stoichiometry to find the other.

Now here we're going to follow the steps to deal with this solutions Stoichiometric question. For step one, it says to convert the given quantity into moles of given. Now if a compound is said to be in excess then just ignore it. So here they're telling us that sodium is an excess, sodium doesn't matter here. Now up above I said that moles equals liters times molarity, so we have to convert our milliliters here into liters. So we have 38.74 milliliters and we're going to stay here for everyone milli it's 10 to the -3 liters.

Now that we have liters, the next thing we can say is that molarity, remember, represents a conversion factor. So that .275 molar of water really means we have .275 moles of water per one liter of solution. So at this point we've just isolated our moles of given. Now Step 2 says we're going to do a mole to mole comparison to convert the given into moles of unknown. So at this point, remember when we do a multiple comparison, we look at the coefficients in the balanced equation and it's two moles of water for every one mole of H₂, which is what we're looking for. So then moles of water down here, moles of H₂ here.

At this point we're done because they're only asking us to find the moles of hydrogen gas. But let's just look at the other steps. If necessary, convert the moles of unknown into the final desired units. O here they only wanted moles of H₂. We can stop. But if they wanted us to go to grams, we'd have to go a little bit further. If they wanted us to find molecules, we'd go a little bit further.

Now Step 4. If you calculate more than one final amount, then you must compare them to determine the theoretical yield. The smaller amount will equal your limiting reagent. So the reactant then makes less product will be the limiting reagent, The one that says it can make more product will be the excess reagent. In this question, we don't have to worry about Step 3 because we only needed moles and we don't have to worry about Step 4 because only one compound had a given amount.

So all we're going to do here is we're going to plug this into our calculator and when we do, we get 5.33×10-3 moles of H2 as our final product for this stoichiometric solution type of question.

How many milliliters of 0.325 M HCl are needed to react with 16.2 g of magnesium metal? 
2 HCl (aq) + Mg (s) → MgCl₂ + H₂ (g)

• A. 33 mL
• B. 742 mL
• C. 872 mL
• D. 3321 mL
• E. 4100 mL

Answer: (E)

What is the molar concentration of a hydrobromic acid solution if it takes 34.12 mL of HBr to completely neutralize 82.56 mL of 0.156 M Ca(OH)­₂? 
2 HBr (aq) + Ca(OH)₂ (aq)  → CaBr₂ (aq) + 2 H₂O (l)

• A. 0.326 M
• B. 0.755 M
• C. 0.811 M
• D. 1.38 M
• E. 1.85 M

Answer: (B)

Consider the following balanced chemical equation: 
H₂O+ 2 MnO₄^– + 3 SO₃^2-  → 2 MnO₂ + 3 SO₄^2-+ 2 OH^–
How many grams of MnO₂ (MW:86.94 g/mol) will be created when 25.0 mL of 0.120 M MnO₄^– (MW:118.90 g/mol) reacts with 32.0 mL of 0.140 M SO₃^2- (MW:80.07 g/mol).  

• A. 0.073 g
• B. 0.089 g
• C. 0.19 g
• D. 0.26 g
• E. 0.67 g

Answer: (D)

Here’s what students ask on this topic:

What is solution stoichiometry and how is it different from regular stoichiometry?

Solution stoichiometry involves stoichiometric calculations in solutions that include volume and molarity, unlike regular stoichiometry which typically deals with masses and moles. In solution stoichiometry, you often start with the volume and molarity of a solution to find the moles of a substance. This is then used in conjunction with a balanced chemical equation to determine the moles of another substance. Finally, these moles can be converted to grams or other units as needed. This approach extends regular stoichiometry by incorporating the concentration of solutions into the calculations.

How do you calculate the grams of a product formed in a solution stoichiometry problem?

To calculate the grams of a product in a solution stoichiometry problem, follow these steps: 1) Convert the given volume to liters if necessary. 2) Use the molarity to find the moles of the given substance using the formula: $\mathrm{moles}=\mathrm{volume}\times \mathrm{molarity}$. 3) Use the mole ratio from the balanced chemical equation to find the moles of the desired product. 4) Convert the moles of the product to grams using its molar mass. This process ensures accurate stoichiometric calculations involving solutions.

What is the role of molarity in solution stoichiometry?

Molarity (M) is a measure of the concentration of a solution, defined as moles of solute per liter of solution. In solution stoichiometry, molarity is crucial because it allows you to convert between the volume of a solution and the moles of solute it contains. This conversion is essential for using the balanced chemical equation to determine the quantities of reactants and products. The formula used is: $\mathrm{moles}=\mathrm{volume}\times \mathrm{molarity}$.

How do you use a balanced chemical equation in solution stoichiometry?

In solution stoichiometry, a balanced chemical equation is used to determine the mole ratio between reactants and products. This ratio is essential for converting the moles of a given substance (found using volume and molarity) to the moles of the desired product. For example, if the balanced equation is $2\mathrm{Na}+2H2O\to H2+2\mathrm{NaOH}$, and you know the moles of $H2O$, you can use the ratio to find the moles of $H2$.

What is the significance of the mole-to-mole comparison in solution stoichiometry?

The mole-to-mole comparison in solution stoichiometry is significant because it allows you to relate the quantities of different substances in a chemical reaction. Using the coefficients from the balanced chemical equation, you can determine how many moles of one substance react with or produce a certain number of moles of another substance. This comparison is essential for converting the moles of a given reactant (found using volume and molarity) to the moles of the desired product, which can then be further converted to grams or other units.