this video will provide a basic introduction into story geometry and for most chemical reactions there's basically three types of conversions that you need to concern yourself with the first type the shortest one is to convert the moles of substance a to the moles of substance B and you need to identify the mol ratio in order to do that number one is going to focus on that problem the second type of problem is to convert the moles of substance a to the G of substance b or you could be given the grams of substance a and you need to convert it to the moles of substance B so that's the second type of problem that you'll see the third type is if you're given the grams of substance a and you need it to convert to the grams of substance B so this involves I believe three steps and this one two steps and the first one is a single step so let's begin working on these problems number one sulfur dioxide reacts with oxygen gas to form sulfur trioxide if 3.4 moles of sulfur dioxide reacts with excess oxygen gas how many moles of sulfur trioxide will form form the first thing that we need to do is we need to write a balanced chemical equation sulur dioxide is SO2 oxygen is diatomic it's O2 and sulfur trioxide is s SO3 so now we need to balance it the sulfur atoms are balanced on both sides but we have four oxygen atoms on the left and three on the right so let's start by putting two in front of2 so now we have two sulfur atoms which means we need to put a two in front of3 now it turns out that the number of oxygen atoms on both sides is now six on this side 2 * 3 is six here 2 * 2 is 4 plus another two that's six so we have a balanced chemical equation at this point now we're given 3.4 moles of sulfur dioxide and we want to convert it to the moles of sulfur trioxide so what we need to do is use something called a mol ratio the M ratio between sulfur dioxide and sulfur trioxide is 2 to2 so what this means is that for every two moles of sulfur dioxide that reacts two moles of sulfur trioxide will be produced now since we have moles of2 on the top left we need to put that on the bottom right so that those units will cancel and the other two moles of SO3 we could put that on top so whenever you want to convert from the moles of substance a to the moles of substance B all you need is one additional fraction beyond what you're starting with and you'll get the answer 2 / 2 is basically one so the answer is 3.4 since the M ratio the coefficients are the same the M ratio is one and so this is the answer for part A based on this example go ahead and try Part B how many moles of oxygen gas will react completely with 4.7 moles of sulfur dioxide so let's start with what we're given which is 4.7 moles of SO2 now we need to convert the moles of SO2 to the moles of oxygen gas so it's a one-step problem all we need is one additional fraction so we need the M ratio between SO2 and O2 so for every two moles of SO2 that reacts one Mo of oxygen gas reacts along with it so let's put the two moles of SO2 on the bottom but on the top we're going to put one mole of o2 based on the coefficient in front of it so make sure you balance the chemical equation before starting this problem so the answer is going to be 4.7 divid 2 which is uh 2.35 moles of o2 as you can see it's not that difficult to convert from the moles of one substance into the moles of another substance number two propane reacts with oxygen gas to form carbon dioxide and water part A if 2.8 moles of propane reacts with excess oxygen gas how many GRS of CO2 will form so let's write down what we need to do we're given the moles of propane C3 h8 and we need to convert it to the gr of carbon dioxide so basically we have the moles of substance a and we want it to convert it to the gr of substance B so it's a two-step process so let's take it one step at a time so starting with the moles of a you want to use the M ratio to change it to a different substance while keeping the unit the same the unit is moles the substance is a so you want to change one thing at a time in this first step we're changing from substance a to substance B now that we have substance B we could change the unit from moles to gr so it's a two-step process for this problem now before we can begin let's write a balanced chemical equation so we have propane C3 h it reacts with oxygen gas to produce carbon dioxide and water so we have three carbon atoms on the left therefore we need to put a three in front of CO2 and there are eight hydrogen atoms on the left and two on the right 2 * 4 is eight so let's put a four in front of water now 3 * 2 is six so we got six oxygen atoms from the 3 CO2 molecules and 4 * 1 is four so we have four oxygen atoms from the four water molecules giving us a total of 10 oxygen atoms on the right side which means we need 10 on the left 10 / 2 is 5 so therefore we need to put a five in front of o2 so now we have a balanced chemical equation now in part A we're given 2.8 moles of propane so first we need to convert or change the substance from propane to carbon dioxide because that's what we're looking for so let's use the M ratio to change the substance so the M ratio between propane and carbon dioxide it's 1 to 3 so for each mole of propane that reacts 3 moles of CO2 will be produced in this reaction so we could cancel these units now we can convert from moles of CO2 to G of CO2 so we need the M mass of carbon dioxide so CO2 contains one carbon and two oxygen atoms so that's 12.01 + 2 * 16 which is going to be 4 24.01 so 1 mole of CO2 has a mass of 44.1 G so whenever you see this number the M Mass the units are G per mole so one mole of that substance has a mass of 44 G so now let's just do the math so we can multiply across it's 2.8 * 3 * 44.0 0 1 and so I got 369.50 gram of carbon dioxide so that's it for part A now let's move on to Part B feel free to pause the video if you want to try it it's very similar to part A how many GRS of oxygen gas will completely react with 3.8 M of propane so we need to convert moles of propane ultimately to G of oxygen so we're going to follow the same steps first we need to change the substance from propane to oxygen but keeping the unit the same so we need to use the M ratio to change a substance after that we could use the M Mass to convert from moles of o2 to G of o2 so that's the blueprint of what we need to follow so let's start with what we're given that is 3.8 moles of propane and let's use the M ratio to convert it to moles of o2 to change a substance so the M ratio is 1 to 5 so one mole of propane C3 h8 will react completely with five moles of o2 so now we're at this point so we need to use the M Mass to go from moles to G now the atomic mass of a single oxygen atom is 16 so the M mass of an oxygen molecule that has two oxygen atoms is 2 * 16 or 32 so so it's 32 G per mole so what that means is that 1 mole of o2 contains a mass of 32 G now we can get the answer so it's 3.8 * 5 * 32 and so it's 68 G of o2 so that's the answer for Part B now let's move on to part C if 25 G of C3 h8 reacts with excess oxygen how many moles of water will form so this time we're given the gr of C3 h8 and we need to convert it to the moles of water so we're given the grams of substance a and we need to convert it to the moles of substance B so first we need to change the unit we need to go from GRS to moles and we're going to use the M Mass to do so and then we're going to use the M ratio to change the substance from A to B so in our particular example we're going to convert the GRS of propane to the moles of propane using the M mass of propane and then we'll use the M ratio to change the substance from propane to water so basically part C is the reverse of A and B so let's begin let's start with what was given to us in the problem that is 25 G of propane now we need to find the M mass of propane it has three carbons and eight hydrogens so that's 3 * 12.01 Plus 8 * 1.8 so that works out to be 44.94 G per mole so one mole of propane has a mass of 44.0 94 grams so we could cancel the unit GRS of C3 H so now let's use the M ratio to change the substance from propane to water so the mol ratio is 1 to 4 so for each mole of propane that reacts four moles of water will be produced now let's perform the operation it's 25 / 44.94 * 4 so this is about 2.27 moles of H2O and that's the answer Part D if 38 G of water are produced in the reaction how many moles of CO2 were produced so this time we're given the gr of water and we need to find the moles of carbon dioxide so just like before first we're going to change the unit from gr to moles using the M Mass and then using the molar ratio we're going to change a substance from water to CO2 so let's begin if you want to pause the video and work on this problem feel free go ahead because the best way to learn is through practice by taking action so let's start with 38 G of H2O and let's convert it to moles so we need to find a m mass of water so we have two hydrogen atoms plus an oxygen atom each hydrogen atom is 1.8 we got to times that by two and then add 16 to it so the M mass of water is 18.016 G per mole so one mole of water has a mass of 18.016 G so now we can move on to our last step and that is converted moles of water to moles of CO2 so the mol ratio is 3 to 4 so for every four moles of water that are produced in this reaction three moles of carbon dioxide are produced along with it so it's going to be 38 / 18.016 multiplied 3 / 4 so the answer that I have have is 1.58 moles of carbon dioxide and that's the final answer number three aluminum reacts with chlorine gas to form aluminum chloride part A if 35 G of aluminum reacts with excess chlorine how many grams of aluminum chloride will form let's start with reaction so we have aluminum reacting with chlorine gas chlorine is diatomic just like oxygen gas and when combined it's going to form aluminum chloride now we need to write the formula of aluminum chloride how can we do so aluminum has a positive3 charge chloride has a minus one charge so using the crisscross method it's going to be al1 cl3 or simply Al cl3 now before we begin the problem we need to balance the chemical equation the number of chlorine atoms is not the same on both sides but the number of aluminum atoms they're equal on both sides so we got two on the left three on the right what I like to do is find the least common multiple of two and three or just multiply two and three which is six so to make them equal I need six chlorine atoms on both sides sides so I'm going to put a three in front of cl2 that's going to give me six and a two in front of al3 so now I have six chlorine atoms but I now have two aluminum atoms on the right side so I got to put a two on the left so now we could focus on part A if 35 gr of aluminum reacts with excess chlorine how many gr of aluminum chloride will form so this is a gram to G conversion we need to convert the gr of substance a to the G of substance B now there are three steps that we need to perform to complete this process first we need to change the units from grams to moles so we need to go from GRS of a to moles of substance a and we need to use the mol mass of substance a to do that next once we have the moles we need to change from substance a to substance B so we need to use the M ratio for that part finally now that we have the moles we need to change the unit from moles to G using the molar mass of substance B so there are three steps that we need to take this is step one step two step three so in our particular example we have the GRS of aluminum and we need to convert it to the moles of aluminum now X we're going to use the M ratio to convert it to the moles of I'm looking at the wrong problem part A we still have the grams of aluminum but we need to convert it to aluminum chloride so we got to get the moles of alcl3 and x and then finally we can convert that to the grams of aluminum chloride so that's an overview of what we need to do in this problem so let's start with what we're given that is 35 G of aluminum based on a periodic table the atomic mass of aluminum is 26.98 so that means 1 Mo of aluminum has a mass of 26.98 G next we need to change the substance from aluminum to aluminum chloride so the M ratio is 2 to2 so for every two moles of Al that reacts 2 moles of aluminum chloride will be produced now the last thing that we need to do is we need to change from the moles of al3 to G so we got to find the M mass of al3 so we need to add the at atomic mass of one aluminum atom with three chlorine atoms so this is 26.98 + 3 times the atomic mass of chlorine which is 35.45 so this is equal to 133 33 G per mole so 133 .33 G of al3 is equivalent to 1 Mo of al3 so now let's do the math it's 35 / 26.98 2 over 2 is 1 so we could cancel the twos if we want to and then let's multiply by 133.33 so the final answer is 17296 G of aluminum chloride so that's the answer for part A Part B how many GRS of chlorine will react completely with 42.8 G of aluminum so feel free to try that problem so we have the grams of aluminum next we need to change it to the moles of aluminum and then we're going to change a substance using M ratio to moles of chlorine and then finally well chlorine diatomic so this is cl2 next we'll change it to the grams of chlorine so that's a gr conversion so let's start with 42.8 g of aluminum and let's convert it to moles using the same atomic mass which is 26.98 now let's use the M ratio to change it from aluminum to cl2 so it's 2: 3 so for every two moles of aluminum that reacts three moles of chlorine gas reacts with it now we need to find the M mass of cl2 it's going to be 35.45 * 2 which is 70.9 so one mole of cl2 is equivalent to 70.9 G so always make sure the other units cancel out if they don't then there's a mistake somewhere so now let's finish the problem it's 42.8 / 26.98 multiplied 3 / 2 and then multiplied by 70.9 so the final answer that I have is 168.75 you know how to perform a mleo mo conversion which we covered in problem one you know how to go from the moles of one substance to the grams of another or the grams of one substance to the moles of a different substance and you know how to perform a gram to gram conversion as well so these are common stochiometry problems that you might see in a typical chemistry class so thanks for watching and have a good day e