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Chemical Reactions and Stoichiometry - Example 3

A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Chemical reactions can be either spontaneous, requiring no input of energy, or non-spontaneous, typically following the input of some type of energy, such as heat, light or electricity. Chemical reactions are usually characterized by a chemical change, and they yield one or more products after the reaction is complete. Chemical reactions are described with chemical equations, which symbolically present the starting materials, end products, and sometimes intermediate products and reaction conditions. Chemical reactions happen at a characteristic reaction rate at a given temperature and chemical concentration. Typically, reaction rates increase with increasing temperature because there is more thermal energy available to reach the activation energy necessary for breaking bonds between atoms.

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

Brown University
Top Chemistry 101 Educators
TD
Theodore D.

Carleton College

Stephen P.

Drexel University

KS
Karli S.

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JH
Jacquelin H.

Brown University