0:00
Hello everyone.
00:01
Our question is it's 22 .7 grams of no, that is nitrogen oxide.
00:09
And 13 .8 gram of o2 are used to form an o2.
00:13
So how many modes of excess reactant will be left over? and the reaction is, the balance reaction i can say is 2 no gaseous plus o2 gaseous, g.
00:28
G2 molds of enno to gaseous.
00:31
Okay.
00:32
So this is the, i can say, balance reaction.
00:36
And we have given that is 22 .7 grams of no is involved and 13 .8 grams of o2 is involved.
00:44
And we have to calculate total number of moles of excess reactants, which will be left over.
00:50
Okay.
00:51
Firstly, we have to calculate the moles, number of moles of each reactor.
00:55
Okay the mass of the en node is given as given masses 22 .7 grams okay and the o2 which is being used or consumed is 13 .8 grams okay now we will calculate the total number of moles of nitrogen oxide an node as well as o2 number of moles are always calculated as mass that is given mass upon molar mass okay upon molar mass of that substance for reactive.
01:30
So, number of moles of n -o is equals to the given mass of no is 22 .7 grams and the molar mass of no is 30 .01 grams per mole.
01:49
That is, i can write it as the addition of mass of gram.
01:56
Molar mass of nitrogen as well as oxysin.
01:59
That is 14 .01 for nitrogen and oxygen consists 16 gram per mole.
02:07
So this is the molar mass of no and after solving this the answer comes out to be that is 0 .7561 moles...