WEBVTT
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for our next problem we have is a octane which
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is a large component of petroleum reacting with oxygen gas
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to produce carbon dioxide. And what? And so
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for the first part of the reaction, we want
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to determine the type of reaction that this is and
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then bounce the equation and then determine the number of
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molecules of carbon dioxide based on the mass given for
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eso opting. So, the first part, we
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know that this type of reaction is a combustion reaction
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because we have our fuel, which is ISO octane
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, reacting with oxygen gas to create the products,
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carbon dioxide and water to turn the balance. Um
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, equation, we can work with one atom at
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a time to make sure we have an equal number
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of atoms on the left and right side for each
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element, and then make sure that thes Doha Kalama
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tree is correct. And so we can start with
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the number off, um, carbon atoms. So
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here we can see that we have eight on our
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left side. So for now we can write a
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coefficient of eight on the right side to make sure
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we have the same number of carbon atoms on both
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sides. Now, if you look at the amount
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of hydrogen, we see that we have 18 on
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the left side, and so we want 18 on
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the right side as well. So for now we
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can write a coefficient of nine in front of water
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. Now, when we look at the number of
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oxygen's on both sides, we see that we have
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to on the left side and a total number of
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16 plus nine, which is 25. And so
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for now we can write, um, 12.5.
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But we know that we can't have half a molecule
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off oxygen, and so we can multiply all of
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this by two. Which means that now we have
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to put a two in front of. I saw
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acting, and we need to put a 25 in
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front of oxygen as well as 16 in front of
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CO two, as well as 18 in front of
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water. And now just a double check our work
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. We see that we have 16 atoms of carbon
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on this side and 16 atoms of carbon on the
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right side, and so that checks out. And
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as for hydrogen, we have a count of 36
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on left side and 36 on the right side.
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And so that checks out. And now, for
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the number of oxygen atoms, we have 50 on
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the left side, and we have 32 plus 18
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, which adds up to 50. And so here
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we can see that we have successfully balanced this equation
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. And now, for the last part of the
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problem, um, if we have 13.2 g of
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iso octane reacting with oxygen cast, um, how
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many molecules of carbon dioxide is produced from this reaction
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? So here we're only given the massive ISO octane
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, so we can assume that the auction gas is
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not a limiting re agent and there is 20.
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So then whenever we calculate number of moles and the
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number of molecules, all of this information will be
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based off of the fact that we have 13.2 g
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of ISO octane. So for this problem, we
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need to use the mass of the ISO octane to
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obtain the moles. Um and then, from there
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we can determine the moles of carbon dioxide and then
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determine the number of molecules using Allah. God rose
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number. So first, let's figure out the molar
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mass of ISO octane. So is so often has
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a chemical formula of C eight, age 18 and
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so we can use this to determine the more mass
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. So we know that there are eight items of
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carbon and so we can multiply the atomic mass of
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carbon by eight and then add this thio eight times
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the atomic mass of Hudgins, which is 1.8 And
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what we're left with is a Mueller Mass of 100
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14.2 g Permal. And now we can use this
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to determine the number of moles of ISO octane that
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are being reacted to in this reaction and so we
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can use dimensional analysis to do this. So we're
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given 13.2 g of ISO octane and we can divide
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this by the more mass obtained from the step before
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to obtain moles of ISO octane. And when we
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do the math, this would mean that we have
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0.11 56 moles of isil octane. And so here
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it's important to note the coefficients because thes actually determined
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, um, the number of moles that you're going
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to work with and so we can use eventual analysis
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to obtain the number of moles of CO two based
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on the number of moles of ISO octane on dso
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. To do this, um, we know that
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we have your point 1156 malls of iso octane and
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so we can multiply this by orisha that we know
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that is trip. So we know that for every
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two moles of I saw octane, we have 16
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goals of co two and we know that this is
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true that this is true because of our balanced equation
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above And so here we can see that there is
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a two in front of the isil octane and we
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have a 16 in front of carbon dioxide. And
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so from there we can write this pre show which
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is completely correct. And from there we can cross
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out the moles of isil often and what we're left
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with is moles of carbon dioxide. And so if
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we do the math for this, we get 0.9247
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moles of co two. So now that we have
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the moles of carbon dioxide, we can figure out
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the number of molecules of carbon dioxide produced from this
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reaction using avocados number. And so we can have
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0.9 few 47 moles of co two You're in here
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and we know that Allah God Rose number is 6.22
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times 10 to the 23rd molecules her one mole of
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substance. And in this case, CO two and
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we're deterring the number of molecules for Coty. And
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so avocados number again is used to convert from units
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of moles, two units of particles. And in
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this case, we're using it to figure out the
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number of molecules for CO two. So when we
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do the math for this, we end up with
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5.569 times, 10 to 23rd molecules of C 02
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And this is written in scientific notation, because this
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is a very large number. And so for this
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problem, we've been able thio identify the type of
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reaction based on the reactant and the products produced,
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and we were able to balance the chemical equation above
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, um, using kind of trial on air and
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accounting for the number of atoms of each element on
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left and right side and for the very last problem
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. We were able to use our knowledge of dimensional
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analysis to figure out the number of moles of ISO
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octane based on the given mass in the problem.
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And we've been able to convert between the moles of
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one molecule to another type of molecule and then use
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of a God rose number to calculate the number of
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molecules that are produced in this reaction.