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Solutions

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

1

Solution Stoichiometry

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solutions to it. Geometry deals with stock geometric calculations in solutions that involve volume and polarity. So if you've watched my videos on stock geometry, you are pretty familiar with our stock geometric chart. Now that we are including solutions toy geometry, we're gonna adapt it slightly to fit. The situations were now volume and polarity are sometimes given. So here, in our solutions Tokyo Metric chart. We're going to say the 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 off water as a liquid, we produce one mole of hydrogen gas and two moles of sodium hydroxide as an acquis product. Now they're giving to us that water has a volume of 38.74 mL off 0.275 Moeller. So this is aren't given information, and then what they're asking us to do is to determine the grams of H two. Now, remember, in our store geometric chart on the left side is where we have our given information. We are cut we're used to seeing are given information given to us in grands or in moles. But now it's solution. Story Geometry will tend to see it also be given to us as volume off Some more clarity. Remember, we've said this before The word off in between two numbers means multiply and realize here that moles equals leaders times more clarity. So if you were to change these MLS toe leaders and multiply them by the mole aren t 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 have given it's our job to get to our unknown are unknown is what we're looking to find here are unknown is h two and we need to find it in grams So following the story geometric chart we have here we go from the moles of given which is H 20 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 multiple comparison where we use the coefficients in the balanced equation. At this point, it would take us two moles of H two and then it's up to us to go to either grams of H two or ions. Adams formula units or molecules of whatever unknown would be. So again, this is a continuation of our idea with Stoke geometry. But now we're including this new shape here. If you haven't watched my video on Stoke geometry, I highly suggest you go back and take a look at the cup of those videos. It's just a continuation of that idea now, including polarity and volume within our calculations. Now that we've seen this new start geometric chart, let's move on to some questions where we put it to practice.

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

2

Solution Stoichiometry Example 1

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in this example question. It says how many moles of hydrogen gas were produced when 38.74 mL off 0.275 more 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 stock geometry, balanced chemical equation. They give us information on at least one compound, and when you start geometry to find the other now here we're gonna follow the steps to deal with this solution. Start geometric 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 leaders times more clarity. So we have to convert our millimeters here into leaders. So we have 38.74 mL and we're going to stay here for everyone. Millie, it's 10 to the negative three leaders. Now that we have leaders, the next thing we can say is that polarity, remember, represents a conversion factor. So that 0.275 Moeller of water really means we have 0.275 moles off water for one leader of solution. So at this point, we've just isolated are moles of Given Now, Step two says we're gonna 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 balance equation, and it's two moles of water for every one mole off H two, which is what we're looking for. So then moles of water down here, Moles of H two 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, the only one that moles of H to weaken stop. But if they wanted us to go to grams, we have to go a little bit further if they wanted us to find molecules to go a little bit further. Now Step four. If you calculate mawr than one final amount, then you must compare them to determinate theoretical yield. The smaller amount will equal your limiting re agent. So the reaction that makes less product will be the limiting regent. The one that says it can make Mawr product will be the excess regent. In this question. We don't have to worry about Step three because we only needed moles. And we don't have to worry about Step four because only one compound how to giving them out. So we're going to do here is we're gonna plug this into our calculator, and when we do, we get 5.33 times 10 to the minus three moles of H two as our final product for this story, you metric solution type of.

3

Problem

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)

A

33 mL

B

742 mL

C

872 mL

D

3321 mL

E

4100 mL

4

Problem

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)

A

0.326 M

B

0.755 M

C

0.811 M

D

1.38 M

E

1.85 M

5

Problem

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).

A

0.073 g

B

0.089 g

C

0.19 g

D

0.26 g

E

0.67 g

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