0:00
Hello everyone.
00:01
Here in this question, it is given that we have 12 .8 grams of nitrogen, 1 .86 gram of hydrogen and 1 .55 gram of ammonia.
00:14
It is shown in the following reaction that nitrogen and hydrogen react to form ammonia.
00:23
And the percentage of this reaction is to be determined in this question.
00:29
From the reaction given here, we can say that one mole of nitrogen react with three more of hydrogen to produce two modes of ammonia.
00:41
Now, let us first determine the limiting reagent involved in this reaction.
00:47
For this, first we have to determine the moles of reactants used in this reaction.
00:54
Let's see how can we do this? for determining the number of moles of reactants, we have to divide the given weight of these reactants and product.
01:06
Let's see how we do it.
01:09
So, number of moles of hydrogen is equal to weight of hydrogen.
01:24
That is given in the question.
01:25
It is 1 .86 divided by its molecular weight.
01:32
Is 2.
01:33
So, the moles of hydrogen used in a reaction is 0 .93.
01:43
Similarly, they calculate the number of moles of nitrogen used in the reaction.
01:54
Number of moles of nitrogen is equal to 12 .8 divided by 28.
02:12
0 .457 modes.
02:16
So here we can observe from this calculation that 0 .457 moles of nitrogen require 0 .93 moles of hydrogen.
02:30
And from this reaction, we know that 1 mole of nitrogen require 3 moles of hydrogen.
02:37
So now let's calculate that how much moles of nitrogen require how much moles of hydrogen in the reaction as one mole nitrogen react with three moles of hydrogen so 0 .457 mole of nitrogen will react with 3 multiplied by 0 .457 moles of hydrogen.
03:23
This comes to be 1 .37 moles of hydrogen.
03:35
So it can be observed from here that for 0 .457 moles of nitrogen is required but in reaction 0 .93 moles of hydrogen is involved.
03:56
And 0 .93 is less than 1 .37 moles of hydrogen...