Gaseous iodine pentafluoride, IF5, can be prepared by the
reacti...

Question

Gaseous iodine pentafluoride, IF5, can be prepared by the 
reaction of solid iodine and gaseous fluorine: 
I21s2 + 5 F21g2¡2 IF51g2 
A 5.00-L flask containing 10.0 g of I2 is charged with 
10.0 g of F2, and the reaction proceeds until one of 
the reagents is completely consumed. After the reaction 
is complete, the temperature in the flask is 125 °C. 
(d) What is the total mass of reactants and products in the flask?

Accepted Answer

Hey everyone in this example, we're told that one of the many interesting features of religions is their ability to form inter halogen compounds. So when bromine reacts with chlorine, bromine, tri chloride is formed in the given reaction below, we're told that in an experiment 12 g of romaine is mixed with 8.50 g of chlorine and the reaction is allowed to continue until complete. We need to calculate the mass of product forms which is our booming tri chloride and determine if any of the reactant are left. If so, we need to calculate how much of any of our reactant are left. So based on the given equation, we can verify that with the given coefficients. This is going to be definitely a balanced reaction which is important to verify. And we want to calculate the theoretical mass of our bro mean tri chloride from our first re agent, which is our romaine gas. And that means we want to look at the ratio between our re Agent bromine gas and our product roaming tri chloride which will come from our balanced equation. And when we look at our balanced equation, we see that we have a ratio or coefficient sorry, of one in front of our bromine gas and a coefficient of two in front of our booming tri chloride, meaning that we have a 1-2 molar ratio. We also want to calculate the theoretical mass of our product. Roaming tri chloride from our second re agent, which is given as chlorine gas. And that means we want to look at the ratio between our re agent chlorine and our product roaming tri chloride and looking at our balanced equation, we have a coefficient of three in front of our chlorine gas and a coefficient of two in front of our booming tri chloride. So we have a 3 to 2 molar ratio. So we're making note of these things because we're going to use them in our calculations. So starting out with the theoretical mass of roaming try chloride from our booming gas, were given a mass of 12 g of grooming and we want to go ahead and convert from grams of bromine, two moles of bromine. And so taking our molar mass from the periodic table for one atom of bromine. We're going to multiply that by two. Since we're dealing with bromine gas, which is going to give us a molar mass equal to 1 59.81 g of bromine for one mole of propane gas. And so now that we have this written out, we want to go from moles of romaine, two moles of our products, bro mean tri chloride. And this is where we're going to plug in that molar ratio where we said we have one mole of bromine gas for two moles of romaine tri chloride that is produced. Next we want to go ahead and move into molds of grooming tri chloride so that we end up with grams of roaming tri chloride And utilizing our molar mass for all of the atoms and roaming tri chloride. We would get from our periodic tables a mass of 186. g for one mole of roaming tri chloride. So now we're able to cancel out our units. We can get rid of moles of chloride and moles of roaming gas as well as grams of roaming gas. Leaving us with grams of roaming tri chloride theoretically produced and this is going to give us a value equal to 28 g of bromine tri chloride that is produced from our first re agent, which is our bromine gas. So moving on to our next theoretically produced mass of roman tri chloride were given a mass for our second reagent of 8.50 g of cl two chlorine gas and we want to go ahead and move from grams of cl 22 moles of cl two just as before. So using our periodic table, we see that for one mole of chlorine, we have 35.45 g. But in this case we're dealing with chlorine gas. So we're going to multiply that by two to get a molar mass equal to 70.91 g for one mole of chlorine gas. Now we want to go ahead and move from moles of chlorine gas, Two moles of grooming tri chloride and then we're going to carry down that molar mass conversion of our bruning tri chloride to cancel out moles of romaine tri chloride. And sorry, this should be moles of B r C L three. So we carried down that same molar mass we said is 1 86.26 g of roaming truck chloride from the periodic table for one mole of romaine tri chloride. And then we plug in that molar ratio of our three moles of chlorine gas which produces two moles of roaming tri chloride, which came from our balanced equation. And so now we can cancel out our units Leaving us with g of roaming tri chloride our product produced. And this is going to give us a value of 14.9 g of roaming tri chloride that is produced. So based on these masses of our product, roaming tri chloride produced, we can deduce that we have or that are 14. g of grooming tri chloride is less than 28 g of roaming tri chloride. And so therefore the re agent that produces this lower mass is going to be The limiting reagent. And we see that we produced 14.9 g of Roman tri chloride from our g of Roman gas. And so we would say that therefore Roman gas is the limiting reagent. And that means it's completely used up in the reaction. And so because we know that bromine gasses are limiting reactant, We can say that therefore cl to our second re agent is left over. And now we're going to have to figure out how much of this reactant is left over. And so to calculate for our mass of chlorine gas left over. And sorry, I almost went too far with this. So this mass of roaming tri chloride was actually produced. As we saw from our chlorine gas here are 8.50 g of chlorine gas produced, the smaller amount of our product. So we would say that the chlorine gasses are limiting reactant and rather our re agent that is left over is going to be our bromine gas. And this also means that the actual amount of of tri chloride that is produced is the amount produced from what we said, our limiting reagent is which is our chlorine gas. So this is the actual amount of grooming tri chloride produced to answer the first part of our question and now to find the second part of our question, we need to see how much gas is left over. However, before we can just find that we need to see how much propane gas is consumed. And so we're going to do this calculation next, Where we take our mass of chlorine gas, which is given as 8.50 g. We're going to go ahead and convert that to moles of cl two. So we're gonna go from g of steel to to moses seal to again, plugging in that molar mass of chlorine gas molecule which from our periodic tables is 70.91 g for one mole. And then we're gonna go ahead and go into the ratio between our limiting reagent chlorine gas and our leftover re agent roaming gas. And from our balanced equation that's going to come from the given reaction above where we have a coefficient of three in front of our chlorine gas and a coefficient of one in front of our bromine gas, which gives us a ratio, Which we can say is a 3-1 ratio here and so we're going to move from moles of cl two are limiting reagent two moles of our excess free agent, the bromine gas that is left over and plugging in that ratio, we had three moles of cl two for one mole of propane gas left over. And then the last part is to get from moles of romaine gas. two g of roaming gas using the molar mass from our periodic table for one mole of propane gas. We would see that for this molecule we have a molar mass of 1 86.26 g of romaine gas. And so now we can go ahead and cancel out moles of roaming gas molds of chlorine gas and grams of chlorine gas to leave us with our grams of our excess re agent or a leftover re agent roaming gas. And this is going to give us a mass of roaming gas that is consumed in this reaction equal to 6. g of our roaming gas consumed in the reaction. And so we can say that therefore we're going to take the difference of the original mass of bromine gas given as 12 g in the problem From our mass of Roman gas consumed, which we just found as 6.38 g. And this difference is going to give us our leftover roaming gas equal to 5.62 g of roaming gas left over. And this is going to be our second and final answer to complete this example. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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