Chemical reactions (stoichiometry)
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Molecular and Empirical Formulas
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The Mole and Avogadro's Number
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Formula from Mass Composition
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Another mass composition problem
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Balancing Chemical Equations
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Balancing Chemical Equations Intuition
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Balancing chemical equations
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Stoichiometry
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Ideal stoichiometry
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Stoichiometry: Limiting Reagent
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Limiting reagent stoichiometry
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Spectrophotometry Introduction
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Spectrophotometry Example
Stoichiometry Introduction to stoichiometry
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- We know what a chemical equation is and we've learned
- how to balance it.
- Now, we're ready to learn about stoichiometry.
- And this is an ultra fancy word that often makes people
- think it's difficult.
- But it's really just the study or the calculation of the
- relationships between the different
- molecules in a reaction.
- This is the actual definition that Wikipedia gives,
- stoichiometry is the calculation of quantitative,
- or measurable, relationships of the
- reactants and the products.
- And you're going to see in chemistry, sometimes people
- use the word reagents.
- For most of our purposes you can use the word reagents and
- reactants interchangeably.
- They're both the reactants in a reaction.
- The reagents are sometimes for special types of reactions
- where you want to throw a reagent in and see if
- something happens.
- And see if your belief about that substance is true or
- things like that.
- But for our purposes a reagent and
- reactant is the same thing.
- So it's a relationship between the reactants and the products
- in a balanced chemical equation.
- So if we're given an unbalanced one, we know how to
- get to the balanced point.
- A balanced chemical equation.
- So let's do some stoichiometry.
- Just so we get practice balancing equations, I'm
- always going to start with unbalanced equations.
- Let's say we have iron three oxide.
- Two iron atoms with three oxygen atoms.
- Plus aluminum, Al.
- And it yields Al2 O3 plus iron.
- So remember when we're doing stoichiometry first of all, we
- want to deal with balanced equations.
- A lot of stoichiometry problems will give you a
- balanced equation.
- But I think it's good practice to actually balance the
- equations ourselves.
- So let's try to balance this one.
- We have two iron atoms here in this iron three oxide.
- How many iron atoms do we have on the right hand side?
- We only have one.
- So let's multiply this by 2 right here.
- All right, oxygen, we have three on this side.
- We have three oxygens on that side.
- That looks good.
- Aluminum, on the left hand side we only have
- one aluminum atom.
- On the right hand side we have two aluminum atoms. So we have
- to put a 2 here.
- And we have balanced this equation.
- So now we're ready to do some stoichiometry.
- There's not just one type of stoichiometry problem, but
- they're all along the lines of, if I give you x grams of
- this how many grams of aluminum do I need to make
- this reaction happen?
- Or if I give you y grams of this molecule and z grams of
- this molecule which one's going to run out first?
- That's all stoichiometry.
- And we'll actually do those exact two types of problems in
- this video.
- So let's say that we were given 85 grams of the iron
- three oxide.
- So my question to you is how many grams of
- aluminum do we need?
- Well you look at the equation, you immediately
- see the mole ratio.
- So for every mole of this, so for every one atom we use of
- iron three oxide we need two aluminums.
- So what we need to do is figure out how many moles of
- this molecule there are in 85 grams. And then we need to
- have twice as many moles of aluminum.
- Because for every mole of the iron three oxide, we have two
- moles of aluminum.
- And we're just looking at the coefficients, we're just
- looking at the numbers.
- One molecule of iron three oxide combines with two
- molecule of aluminum to make this reaction happen.
- So lets first figure out how many moles 85 grams are.
- So what's the atomic mass or the mass number
- of this entire molecule?
- Let me do it down here.
- So we have two irons and three oxygens.
- So let me go down and figure out the atomic masses of iron
- and oxygen.
- So iron is right here, 55.85.
- I think it's fair enough to round to 56.
- Let's say we're dealing with the version of iron, the
- isotope of iron, that has 30 neutrons.
- So it has an atomic mass number of 56.
- So iron has 56 atomic mass number.
- And then oxygen, we already know, is 16.
- Iron was 56.
- This mass is going to be 2 times 56 plus 3 times 16.
- We can do that in our heads.
- But this isn't a math video, so I'll get
- the calculator out.
- 2 times 56 plus 3 times 16 is equal to 160.
- Is that right?
- That's 48 plus 112, right, 160.
- So one molecule of iron three oxide is going to be 160
- atomic mass units.
- So one mole or 6.02 times 10 to the 23 molecules of iron
- oxide is going to have a mass of 160 grams.
- So in our reaction we said we're starting off with 85
- grams of iron oxide.
- How many moles is that?
- Well 85 grams of iron three oxide is equal
- to 85 over 160 moles.
- So that's equal to, 85 divided by 160 equals 0.53125.
- Equals 0.53 moles.
- So everything we've done so far in this green and light
- blue, we figured out how many moles 85 grams of
- iron three oxide is.
- And we figured out it's 0.53 moles.
- Because a full mole would have been 160 grams. But
- we only have 85.
- So it's point 0.53 moles.
- And we know from this balanced equation, that for every mole
- of iron three oxide we have, we need to have
- two moles of aluminum.
- So if we have 0.53 moles of the iron molecule, iron three
- oxide, then we're going to need twice as many aluminum.
- So we're going to need 1.06 moles of aluminum.
- I just took 0.53 times 2.
- Because the ratio is 1:2.
- For every molecule of this, we need two molecules of that.
- So for every mole of this, we need two moles of this.
- If we have 0.53 moles, you multiply that by 2, and you
- have 1.06 moles of aluminum.
- All right, so we just have to figure out how many grams is a
- mole of aluminum and then multiply that times 1.06 and
- we're done.
- So aluminum, or aluminium as some of our friends across the
- pond might say.
- Aluminium, actually I enjoy that more.
- Aluminium has the atomic weight or the
- weighted average is 26.98.
- But let's just say that the aluminium that we're dealing
- with has a mass of 27 atomic mass units.
- So one aluminum is 27 atomic mass units.
- So one mole of aluminium is going to be 27 grams. Or 6.02
- times 10 to 23 aluminium atoms is going to be 27 grams. So if
- we need 1.06 moles, how many is that going to be?
- So 1.06 moles of aluminium is equal to 1.06 times 27 grams.
- And what is that?
- 1.06 times 27.
- Equals 28.62.
- So we need 28.62 grams of aluminium, I won't write the
- whole thing there, in order to essentially use up our 85
- grams of the iron three oxide.
- And if we had more than 28.62 grams of aluminium, then
- they'll be left over after this reaction happens.
- Assuming we keep mixing it nicely and the whole reaction
- happens all the way.
- And we'll talk more about that in the future.
- And in that situation where we have more than 28.63 grams of
- aluminium, then this molecule will be the limiting reagent.
- Because we had more than enough of this, so this is
- what's going to limit the amount of this
- process from happening.
- If we have less than 28.63 grams of, I'll start saying
- aluminum, then the aluminum will be the limiting reagent,
- because then we wouldn't be able to use all the 85 grams
- of our iron molecule, or our iron three oxide molecule.
- Anyway, I don't want to confuse you in the end with
- that limiting reagents.
- In the next video, we'll do a whole problem devoted to
- limiting reagents.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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