Solution stoichiometry deals with stoichiometric calculations in solutions that involve volume and molarity. 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 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 moles of sodium solid reacting with 2 moles of water as a liquid, we produce 1 mole of hydrogen gas and 2 moles of sodium hydroxide as an aqueous product. It is given to us 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 H_{2}. In our stoichiometric chart, on the left side, we have our given information. We are used to seeing our given information in grams or in moles, but now with solution stoichiometry, it will also be given to us as a volume of some molarity. Remember, we've said this before, the word "of" between two numbers means multiply. Realize here that moles equal liters times molarity. If you were to convert these milliliters 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. Once we have our moles of given, it's our job to get to our unknown. Our unknown here is H_{2}, and we need to find it in grams. Following the stoichiometric chart here, we go from the moles of given, which is H_{2}O, 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. 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 H_{2}, and then it's up to us to go to either grams of H_{2}, or ions, atoms, formula units, or molecules of whatever unknown would be. This is a continuation of our idea with stoichiometry but now includes 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.

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# 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: ${\mathrm{moles}}^{=}\mathrm{liters}\mathrm{molarity}$. This process allows for mole-to-mole comparisons to find unknowns, such as grams of hydrogen gas produced in a reaction. Understanding this concept is essential for mastering stoichiometric calculations in chemistry.

**Solution Stoichiometry** deals with stoichiometric calculations in solutions that involve volume and molarity.

## Solution Stoichiometry

### Solution Stoichiometry

#### Video transcript

Use the solution stoichiometric chart when dealing with calculations involving molarity and/or volume.

### Solution Stoichiometry Example 1

#### Video transcript

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. Alright. So we know that this is a balanced chemical equation. They are giving us information on 1 compound and asking for information on another. This is the definition of stoichiometry. The chemical equation gives us information on at least 1 compound, and we use stoichiometry to find the other. Now here we're going to follow the steps to deal with this solution stoichiometric question. For step 1, it says to convert the given quantity into moles of given. 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. We have to convert our milliliters here into liters. 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 step is to acknowledge that molarity 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. 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. At this point, 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 H₂, which is what we're looking for. 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 mole units. Here they only wanted moles of H₂ so 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 1 compound had a given amount. So all we're going to do here is plug this into our calculator and when we do, we get \(5.33 \times 10^{-3}\) moles of H₂ 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) → MgCl_{2} + H_{2} (g)

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) → CaBr_{2} (aq) + 2 H_{2}O (l)

Consider the following balanced chemical equation:

H_{2}O+ 2 MnO_{4}^{–} + 3 SO_{3}^{2-} → 2 MnO_{2} + 3 SO_{4}^{2-}+ 2 OH^{–}

How many grams of MnO_{2} (MW:86.94 g/mol) will be created when 25.0 mL of 0.120 M MnO_{4}^{–} (MW:118.90 g/mol) reacts with 32.0 mL of 0.140 M SO_{3}^{2-} (MW:80.07 g/mol).

### Here’s what students ask on this topic:

What is solution stoichiometry and how is it used in chemical reactions?

Solution stoichiometry involves using the volume and molarity of solutions to determine unknown quantities in chemical reactions. It extends traditional stoichiometry by incorporating the relationship between moles, volume, and molarity. For example, given a balanced chemical equation, you can convert the volume of a solution (in mL) and its molarity to moles using the equation:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

This allows for mole-to-mole comparisons to find unknowns, such as the grams of a product formed. Understanding this concept is crucial for mastering stoichiometric calculations in chemistry.

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:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

First, convert the volume from milliliters (mL) to liters (L) by dividing by 1000. Then, multiply the volume in liters by the molarity (M) of the solution. 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:

$0.038740.275=0.010654\mathrm{moles}$

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

To perform a mole-to-mole comparison in solution stoichiometry, use the coefficients from the balanced chemical equation. For example, consider the reaction:

$2\mathrm{Na}\left(s\right)+2H2O\left(l\right)\to H2\left(g\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)$

If you know the moles of H_{2}O, you can find the moles of H_{2} produced by using the ratio from the balanced equation. Here, 2 moles of H_{2}O produce 1 mole of H_{2}. So, if you have 0.010654 moles of H_{2}O, the moles of H_{2} produced would be:

$\frac{0.010654}{2}=0.005327\mathrm{moles}H2$

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

To calculate the grams of a product formed in a solution stoichiometry problem, follow these steps:

1. Convert the volume and molarity of the reactant to moles using:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

2. Use the balanced chemical equation to perform a mole-to-mole comparison.

3. Convert the moles of the product to grams using its molar mass. For example, if you have 0.005327 moles of H_{2} and the molar mass of H_{2} is 2.02 g/mol, the mass in grams is:

$0.0053272.02=0.01076\mathrm{grams}$

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

The relationship between molarity, volume, and moles in solution stoichiometry is given by the equation:

$\mathrm{moles}=\mathrm{liters}\mathrm{molarity}$

Molarity (M) is defined as the number of moles of solute per liter of solution. To find the moles of solute, multiply the volume of the solution (in liters) by its molarity. For example, if you have 0.5 L of a 1.5 M solution, the moles of solute are:

$0.51.5=0.75\mathrm{moles}$