When benzene 1C6H62 reacts with bromine 1Br22, bromobenzene 1C6H5Br2 is obtained: C6H6 + Br2¡C6H5Br + HBr (a) When 30.0 g of benzene reacts with 65.0 g of bromine, what is the theoretical yield of bromobenzene?

Video transcript

hey everyone in this example, we have ethanol reacting with gaseous hydrogen chloride to form chloral methane and water. So it's given in the reaction below and we need to figure out how much chloral ethane our product can be produced from the reaction if 15 g of ethanol and 20 g of hydrogen chloride are our re agents used. So this is essentially asking us for our theoretical yield of our product here. And so what we want to do is find how much chloral ethane. So C two H five cl produced from our first re agent, which is our ethanol. In our first part of our calculation. And this is again going to be a theoretical calculation. And then we want to also do another theoretical calculation finding the mass of our chloral ethane produced from our second re agent hydrochloric acid. And so we want to make note of the ratios between our agent and the product. So looking at the ratio between our ethanol and our chloral ethane, we're going to get that from our given reaction. We want to make sure that this is a balanced reaction. And based on all of our coefficients in this given reaction, we can confirm that this is balanced, which is important because we're going to use these coefficients to determine this ratio. And so looking at our Ethan, all we would see that we have a coefficient of one. And looking at our chloral ethane, we have a coefficient of one in front of that too. And so we have therefore a 1-1 molar ratio. We also want to make note of the molar ratio between our chloral ethane and our re agent hydrochloric acid. And rather I'm just going to write it in the same order. So a ratio between our re agent hydrochloric acid and our product chloral ethane. And so we would see in our reaction, we also have a coefficient of one in front of our hcl. So we have a 1-1 molar ratio here as well because we still have that coefficient in front of arklow ethane of one. So now we can get into the calculations and were given a mass four hour ethanol of 15 g. We want to go ahead and convert from grams of ethanol into molds of ethanol. And so using the periodic table, we're going to find the molar mass for every atom within ethanol equal to a value of 46.7 g for one mole next, we want to go ahead and move from moles of ethanol into molds of our product chloral ethane. And then lastly, we want to go in from molds of chloral ethane, two g of chloral ethane. So what we're going to do is for the molar ratio portion, we want to refer to the ratio that we took note of here and we said that we have from one mole of our ethanol, we produce one mole of chloral methane. And lastly we want to plug in that molar mass of our chloral ethane from our periodic table, which we would see is equal to a value of 65 or 64.51 g of methane for one mole of methane. Now we're able to cancel out our units so we can get rid of moles of methane moles of ethanol and grams of ethanol leaving us with grams of chloral methane produced. And this is going to give us a value equal to 21.0 g of chloral ethane that is produced Theoretically. So, moving on to the second part of our calculation were given a mass of 20 g of hcl. And so we want to go ahead and move from grams of hcl two moles of hcl. So using the molar mass in the periodic table, we would see that for one mole of hcl we have a molar mass of 36.46 g for every atom in hcl. And now we're going to multiply so that we can grow from moles of hcl, two moles of our product, Claure thane. Sorry, that should be a two there. And lastly we want to go from molds of chloral ethane, two moles of our product or sorry, two g of our product. Urethane, just like before and again, we would plug in that same molar mass, 64.51 from the periodic table for chloral ethane grams for one mole of methane And then we need to plug in that molar ratio. So we said that from our balanced equation, we have one mole of hydrochloric acid producing one mole of our chloral ethane. And so now we can go ahead and cancel out our units, Leaving us with g of chloral ethane produced. And this is going to give us a value equal to 35.4 g of chloral ethane that is produced from our second re agent hydrochloric acid. And so what we can say is that 21 g of chloral ethane Is less than 35. g of chlorine methane. And so therefore the re agent that produces this lower mass which was produced from our ethanol here or 15 g of ethanol is going to be our limiting re agent. And so we would say that ethanol is our limiting re agent. And so because we know that that's our limiting reagent. Thus we can say 21.0 g of our chloral ethane is produced and this is going to be our final answer to complete this example. So I hope that everything I reviewed was clear. But if you have any questions, please leave them down below. And I will see everyone in the next practice video