00:01
Hello students for the given reaction, we need to tell the limiting reactant mass of no and h2o and the reactant remain after limiting reactant has reacted.
00:11
So for this we have given mass of ammonia and oxygen.
00:15
So first we will calculate the number of moles which will be equals to given mass divided by molar mass.
00:26
So moles of ammonia will be equals to 1 .65 divided by 17 which is equals to 0 .09 mole and moles of o2 will be equals to 2 .15 divided by 32 which will be equals to 0 .06 mole.
00:56
So from here we can see that o2 is the limiting reactant.
01:10
For the next part we need to calculate the mass of no.
01:19
So from reaction we can see that 5 mole of o2 will produce 4 mole of no.
01:30
So 1 mole will give 4 by 5 mole that means 0 .06 moles will give 4 by 5 multiplied by 0 .06 which will be equals to 0 .048 mole of no.
01:53
So mass will be equals to number of moles multiplied by molecular mass.
01:58
So this will be equals to 0 .048 multiplied by molecular mass of no is 30.
02:07
So this will be equals to 1 .4 gram...