00:01
Hello student, so the question is to calculate the mass of the precipitate which is crpo4 formed in this reaction.
00:09
So in step 1, we will be calculating the number of moles of the reactants individually.
00:17
So here first we will calculate the number of moles of the na3po4.
00:29
So here we will be using this formula where the number of moles denoted by n is equal to the molarity into the volume in liters.
00:40
So here we have been given the concentration of the na3po4 which is, so here the concentration of the na3po4 is equal to 0 .3 molar and the volume of na3po4 is given as 45 .5 ml.
01:05
So we need to convert it into liters, so we will divide it by 1000.
01:10
So it is equal to 0 .0455 liter.
01:15
So now let's put the value over here to calculate the number of moles of the na3po4 which is equal to, so let's put the value, so the molarity is 0 .3 molar multiplied by the volume which is 0 .0455 liter.
01:33
So here it is equal to, which is equal to 0 .01365 moles.
01:42
So this is the number of moles of na3po4.
01:46
Now similarly we will be calculating the number of moles of the crcl3.
01:53
So here we have been given the molarity which is the concentration of the crcl3 which is equal to 0 .2 molar and the volume is given as, it is given as 4 point, which is given as 44 ml.
02:13
So in liters it will be equal to 0 .044 liter.
02:18
So now let's put the value over here in this formula.
02:21
So here the molarity is 0 .2 molar multiplied by the volume which is 0 .044 liter.
02:28
So by calculating it we will get the number of moles and it is equal to 0 .088 moles.
02:35
So this is the number of moles of crcl3.
02:38
Now we will need to find the limiting reagent.
02:41
Now the step 2 is to find the limiting reagent.
02:53
So here as from the equation given here we can see that the reactant, the 1 mole of na3po4 is been using, is been reacted with 1 mole of the crcl3.
03:17
So here the mole ratio is of 1 is to 1...