Solution Stoichiometry - Online Tutor, Practice Problems & Exam Prep

Solution stoichiometry involves calculations using volume and molarity to determine unknown quantities in chemical reactions. For example, using a balanced equation, one can convert the volume of a solution (in mL) and its molarity to moles, applying the relationship: moles = liters × molarity. This allows for mole-to-mole comparisons based on coefficients in the balanced equation, ultimately leading to the calculation of grams of a product, such as hydrogen gas (H₂). Understanding this process is essential for mastering stoichiometric calculations in solutions.

Solution Stoichiometry

Solution 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 solution stoichiometry, we're going to adapt it slightly to fit the situations where now volume and molarity are sometimes given. Here in our solution stoichiometric chart, we're going to say a chart uses the given quantity of a compound to determine the unknown quantity of another compound.

In this example, we have our balanced chemical equation which states: for every 2 Na(s) reacting with 2 H2 O (l), we produce 1 H2 (g) and 2 NaOH (aq). We are given that water has a volume of 38.74 milliliters of 0.275 molar. This is our given information. We are asked to determine the grams of H₂.

Remember in our stoichiometric chart, on the left side is where we have our given information. We are used to seeing our given information in grams or moles, but now with solution stoichiometry, it will also be given as a volume of some molarity. Recall that "of" between two numbers means multiply. Note here that moles equal liters times molarity. So, if you were to convert these mL to liters and multiply them by the molarity, then this new given shape we are seeing feeds directly into our moles of given.

Once we have our moles of given, it's our job to get to our unknown. Our unknown is what we are looking to find. Here our unknown is H₂ and we need to find it in grams. Using the stoichiometric chart, we go from the moles of given, which is H₂O, and we make the jump. In this jump, we're moving from an area where we have information we know to where we don't have any information about the unknown. This requires a mole to mole comparison, using 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 the unknown would be.

This method is a continuation of our understanding with stoichiometry, but now we include molarity and volume in our calculations. If you haven't watched my video on stoichiometry, I highly suggest you go back and check out a couple of those videos. We are moving forward with this idea, now incorporating volume and molarity into 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 1

In this example question, it asks how many moles of hydrogen gas were produced when 38.74 milliliters of 0.275 molar water react with excess sodium. Alright. So we know that this is a balanced chemical equation. They're providing information on one compound and asking for information on another. This is the definition of stoichiometry. The chemical equation gives 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 stoichiometric question.

For step 1, it says to convert the given quantity into moles of the given compound. If a compound is said to be in excess, then just ignore it. So here, they're telling us that sodium is in excess, so 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 say here for every 1 milliliter it's \(10^{-3}\) liters. Now that we have liters, the next thing we can say is that molarity, remember, represents a conversion factor. So that 0.275 molar of water really means we have 0.275 moles of water per 1 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 the unknown. So at this point, remember when we do a mole-to-mole comparison, we look at the coefficients in the balanced equation. It's 2 moles of water for every 1 mole of \(H_2\), which is what we're looking for. So then, moles of water down here, moles of \(H_2\) 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 mole units. So here they only wanted moles of \(H_2\) 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 that 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 \(5.33 \times 10^{-3}\) moles of \(H_2\) as our final product for this stoichiometric solution type.

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

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

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

Here’s what students ask on this topic:

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

Solution stoichiometry involves calculations using the volume and molarity of solutions to determine unknown quantities in chemical reactions. Unlike regular stoichiometry, which typically deals with masses and moles of reactants and products, solution stoichiometry incorporates the concentration of solutions. The key relationship used is: $\mathrm{moles}=\mathrm{liters}\times \mathrm{molarity}$. This allows for mole-to-mole comparisons based on coefficients in the balanced equation, ultimately leading to the calculation of quantities like grams of a product.

How do you convert volume and molarity to moles in solution stoichiometry?

To convert volume and molarity to moles in solution stoichiometry, use the equation: $\mathrm{moles}=\mathrm{liters}\times \mathrm{molarity}$. First, convert the volume from milliliters to liters by dividing by 1000. Then, multiply the volume in liters by the molarity (moles per liter) to find the number of moles. For example, if you have 38.74 mL of a 0.275 M solution, convert 38.74 mL to 0.03874 L and then multiply by 0.275 M to get the moles.

How do you perform a mole-to-mole comparison in solution stoichiometry?

In solution stoichiometry, a mole-to-mole comparison is performed using the coefficients from the balanced chemical equation. After converting the given volume and molarity to moles, use the coefficients to relate the moles of the given substance to the moles of the unknown substance. For example, in the reaction $2\mathrm{Na}+2H2O\to H2+2\mathrm{NaOH}$, if you know the moles of $H2O$, you can find the moles of $H2$ by using the ratio $\frac{1}{2}$.

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

To calculate the grams of a product in solution stoichiometry, follow these steps: 1) Convert the given volume and molarity to moles using $\mathrm{moles}=\mathrm{liters}\times \mathrm{molarity}$. 2) Perform a mole-to-mole comparison using the balanced chemical equation. 3) Convert the moles of the product to grams using its molar mass. For example, if you need to find the grams of $H2$ produced, first find the moles of $H2$ and then multiply by the molar mass of $H2$ (2.02 g/mol).

What is the relationship between molarity and volume in solution stoichiometry?

In solution stoichiometry, the relationship between molarity and volume is crucial for converting between moles and the volume of a solution. The key equation is: $\mathrm{moles}=\mathrm{liters}\times \mathrm{molarity}$. This means that the number of moles of a solute in a solution is equal to the volume of the solution (in liters) multiplied by its molarity (moles per liter). This relationship allows you to determine the amount of substance present in a given volume of solution, which is essential for stoichiometric calculations.