00:01
Hello students let's begin with this question.
00:02
This question is based upon the saponification of the trimstine and some data is provided in the question.
00:13
We have to find out the amount of the naoh that is required to react with the given amount of the trimstine.
00:19
Now firstly the mass of the trimstine is given in the question.
00:26
So this value is equals to 579 gram and the molar mass of the trimstine is given.
00:37
So firstly its formula is also given in the question.
00:41
The trimstine's formula is it is composed of carbon, hydrogen and oxygen.
00:47
So here its formula is c45h86o6.
00:56
Its molar mass is equals to 723 .16 gram per mole.
01:03
Now we have the mass and the molar mass of the trimstine.
01:06
From this by using the formula that moles is equals to given mass upon the molar mass of the substance we can find out the moles of the trimstine.
01:19
So here the moles of trimstine comes to be this is equals to its mass is 579 gram divided by its molar mass is 723 .16 gram per mole.
01:36
So when we calculate this we get the moles of the trimstine.
01:40
So its moles comes to be 0 .8006 moles.
01:45
Now from the balanced equation from this equation it is clear that the one mole of trimstine it will react with the three moles of the noh...