00:01
Hello students today we have been provided with the question having four subpart so let's take up the first part and the equation that has been provided to us is p4 plus 6 cl2 gives 4 pcl3 that is your phosphorus trichloride okay so the very first thing that we are going to do over here is to calculate the number of mole of substance so the number of modes of p4 we are going to calculate first of all so write down number of more of more of p4 we have the formula for calculating this that is mass of the substance so here mass of p4 mass of p4 divided by molar mass of molar mass of molar mass of p4 okay so putting the values we'll get 20 .00 gram divided by 123 .895 grams per mole cut grams to gram you are going to get 0 .161 mole okay so this is the number of moles of p4 right so now what you can see from this particular equation take it as one see one of p4 react with six moles of cl2 write down one mole of p4 reacts with reacts with six moles of cl2 right so point 61 so point 161 mole of p4 will react with 6 multiplied by 0 .161 mole of co2 right so that would be 0 .966 moles when you multiply 0 .161 with 6 you'll get 0 .966 moles okay so like that we'll calculate the mass of chlorine as well therefore therefore mass of cl2 that would be equal to number of moles of cl2 multiplied by molar mass of cl2 okay molar mass of cl2 right and that would be 0 .966 mole multiplied by molar mass of cl2 would be 70 .95 gram per mole.
02:34
Cut mole with mole you will get over here 68 .4894 grams.
02:44
So that is the first part ended over here where you can conclude that 68 .4894 gram of cl2 are needed to react.
02:54
So now let's switch to our b part.
02:57
Okay.
02:58
So now we'll again calculate the number.
03:00
Of moles of p4 number of number of moles of p4 the formula will remain same molar mass sorry mass of p4 upon molar mass so mass over here is different from the a part here it is 15 .00 grams divided by 123 .895 gram per mole cut grams to gram you will get 0 .1 to 1 .1 more okay and also calculate the number of moles of cl2 number of moles of cl2 that would be again mass upon molar mass that would be 22 .00 gram provided in the position divisible by 70 .1990 gram per mole and cut gram to gram you'll obtain over here is 0 .310 moles.
03:55
All right.
03:56
So so now now, from equation 1, what you can say that 1 mole, 1 mole of p4, 1 mole of p4 reacts with 6 chlorine, right? 6 moles of chlorine, 6 moles of chlorine, okay? while 1 .1 to 1 mole of p4 react with 6 multiplied by 0 .1 to 1 moles of chlorine, right? so that would be equal to 0 .726 moles.
04:36
Okay, that is one thing, that is one result that we have obtained over here.
04:41
Now, what we have to see is, see, we have only 0 .310 moles of cl2 over here, you can see, right? this is, you can see.
04:52
We have only 0 .310 modes of chlorine and we need at least 0 .726 moles.
04:58
So, ultimately what you can see over here that chlorine is the limiting reactant, okay? so, cl2 is going to be your limiting reactant, right? and, and p4 is going to be your excess reactant.
05:21
That is one thing.
05:23
Okay? so, okay, let's just switch to the next slide.
05:26
So as for the same reaction, what we can say that 6 moles, the equation 1 only, 6 moles of chlorine react with 1 mole of p4, 1 mole of p4, while 1 mole of chlorine react with 1 by 6 mole of p4, right? then after what we can say 0 .310 moles of cl2 react with 1 by 6 multiplied by 1 divided by 6 multiplied by 3 .310 moles of moles of p 4 which is equal to 0 .051 moles okay right so therefore therefore p 4 remains therefore 3 4 remains is equal to 0 .1 .1 minus 0 .051 equal to 0 .07 moles.
06:35
Okay, this much of p4 remains.
06:37
Now again from equation 1 what you can see? we can see right down from same equation 1, we can say that 6 moles, 6 moles of chlorine, 6 .4.
06:53
Moles of chlorine gives gives four moles of pcl3 okay so one mole that is one okay one mole of cl2 gives 4 by 6 mole of pcl 3 so 0 .310 mole of cl2 will give 4 by 6 multiplied by 4 divided by 6 multiplied by 0 .3 .0.
07:23
0 .310.
07:26
Let me do it on the upper side so that that would be more clear to you.
07:29
Alright, so that i have switched to the upper side.
07:32
So you can see 6 moles of chlorine gives 4 moles of pcl3 and 1 more will give 4 by 6 more of tcl3.
07:39
Then 0 .310 moles of cl2 give 4 by 6 multiplied by 0 .310 moles of pcl3.
07:45
So that is going to be 0 .206 modes, okay? so, in that case, therefore, therefore, mass of pcl3 would be equal to number of moles of pcl3 multiplied by molar mass.
08:02
So i'm writing the values directly, 0 .206 moles, multiplied by 137, that is a molar mass of pcl3, 137 .33 gram per moles.
08:15
Okay, you can cut moles to moles and you'll get the 1 .307.
08:17
Value that is 28 .28 grams right so therefore you can say that 28 .28 gram of pcl 3 are produced okay it is not over yet b portion is not over 8 so let me switch to the next slide again now number of number of moles of excess number of excess reactant number of excess reactant p4 remains after completion that would be 0 .07 moles okay so therefore mass of p4 mass of p4 is equals to again number of moles of p4 multiple by molar mass of p4 that is 0 .7 moles multiplied by 123 .895 grams per mole okay so you can cut more to more and you will get eventually your mass of p4 that is 8 .672 grams so here you can come you are you can conclude your b part of this question that is excess excess reactant p4 remains, excess reactant p4 remains would be in grams equal to 8 .672 grams.
09:50
So now we'll be doing our c part.
09:53
So let's do it.
09:54
So in c part we have to calculate the percentage yield.
09:58
So to calculate the percentage yield, we have the formula.
10:01
Actual yield divisible by theoretical yield multiplied by 100%.
10:06
Okay.
10:07
Actual yield, divisible by theoretical yield, theoretical yield multiplied by 100%.
10:15
So in the question we have been provided with these two values, actual as well as theoretical yield.
10:21
So i'm going to put it over here directly...