00:01
Hello everyone.
00:02
So in this question we have been given with a reaction where aluminium iodide is reacting with mercury chloride giving rise to aluminium chloride and mercury iodide and we have been asked here to find out the aluminium iodide consumed in order to prepare 46 grams of mercury iodide here.
00:24
So we have been given with the reaction where aluminium iodide we have been given with a balance chemical.
00:35
Reaction.
00:37
So 2 moles of 2 moles of aluminium iodide reacting with 3 moles of mercury chloride giving rise to 2 moles of aluminium chloride and 3 moles of mercury iodide.
01:09
So we have to find out the aluminium iodide consumed here in order to prepare 46 grams of mercury iodide.
01:23
So 1 mole of mercury iodide that is hg i2 it weighs for mercury that is 200 .5 gram while for 1 iodine it is 127 while we have we have i2 here that means 2 multiplied by 127 which will which will be eventually equal to 4 24 .5.
02:09
2 multiplied by 127 that will be equal to 454 .5 gram.
02:19
So that's the molecular weight of 1 mole of mercury iodide while we have 3 moles of it.
02:26
So 3 moles of mercury iodide will weigh 3 multiplied by 454 .5 grams.
02:45
So that's the molecular weight of 3 moles of mercury iodide.
02:54
Now let's move on to the molecular weight of aluminium iodide.
02:58
So 1 mole of aluminium iodide, that is ali3, it weighs for aluminium it is 27 gram plus 3 multiplied by 127, which will be equal to 408 gram.
03:21
So that's the molecular weight of 1 mole of aluminium iodide.
03:24
While in the chemical reaction we have 2 moles of it.
03:31
So we can say that 3 multiplied by 454 .5 gram of mercury iodide that hgi .2 is obtained from 2 moles of aluminum iodide.
04:01
That means 2 multiplied by 408 gram of aluminum iodide.
04:19
So 1 gram of it will be obtained from 1 gram of mercury iodide will be obtained from 2 multiplied by 408 divided by 3 multiplied by 450 4 .5 grams of aluminum iodide.
04:49
Well according to the equation we have 46 grams of mercury iodide so for so in order to prepare 46 gram of mercury iodide for 46 grams of mercury iodide the amount of or the mass of aluminium iodide that we require will be equal to 2 multiplied by 408 divided by 3 multiplied by 454 .5 multiplied by 46 that will be equal to 27 .53 grams.
05:32
Aluminium iodide required in order to prepare 46 grams of mercury iodide will be equal to 27 .53 grams.
05:44
So that's the answer of our first question.
05:47
Now let's move on to the second one.
05:52
Now the second question says we have been given with the balanced chemical reaction where 3 moles of hydrogen reacting with 1 moles of nitrogen giving less to 2 moles of ammonia we have been asked to find out the mass of ammonia obtained in grams when 77 gram of nitrogen is consumed so we have the reaction 3 moles of nitrogen reacting with 3 moles of hydrogen reacting with 1 mole of nitrogen giving rise to 2 moles of ammonia and we have the and we have to find out amount of product formed in grams when 77 gram of nitrogen is consumed so we can see from here in the chemical reaction that 1 mole of nitrogen giving rise to 2 moles of ammonia okay so 1 mole of nitrogen giving rise to 2 moles of ammonia and and 1 mole of nitrogen weighs 28 gram while 1 mole of ammonia weighs 17 gram.
07:23
How? for nitrogen it is 14 while for hydrogen 3 multiplied by 1.
07:28
So we can say that 1 mole of ammonia weighs 17 gram while we have 2 more of it.
07:33
So we can say that 28 gram of nitrogen here is giving rise to...