00:01
Hello students welcome here in this question there are two parts so the first part is reaction of magnesium acetate is given so ch3 c -o -o -m -z so this one is two times okay so this one is a case solution and it is reacted with sodium sulphide and sodium sulphide means n -a -2 s actus so it gives reaction magnesium sulfide is formed and 2 moles of ch3 c -o -o -n -a is produced so this one is solid and this one is active solution so we have to calculate the mass of this one when the compound so here it is excess magnesium estate is taken this is in excess and this one is 15m.
01:05
55 .2ml.
01:07
0 .0 .0 .204m solution and a2 sodium sulfide is taken.
01:14
So, and excess is magnesium acetate.
01:21
Now, we have to calculate the mass.
01:23
So here we can see the reaction is already balanced.
01:30
So one more of n .a2 s is giving one mole of mgs.
01:39
So how many moles here it will be produced 55 .02 ml multiplied by 0 .2045m means.
01:52
Here how much mass is present here? it is multiplied by 10 power minus 3 because it is in ml.
02:00
But we want liters.
02:02
So that's why here it is 10 .5 minus 3.
02:05
So number of moles is equal to 0 .0113 moles of n .a .2 is reacting...