00:01
Let's discuss this question.
00:03
So here it is given to us that dichloromethane is formed from methane and chlorine and here the mass of methane is given as 1500 grams and percentage of the reaction is 82 .5%.
00:17
So firstly here we need to find the balance reaction that is we need to write the balance equation.
00:31
Secondly, here we need to find the mass of chlorine required.
00:43
So let's start here we can say the reaction will be when ch4 methane reacts with cl2, that is chlorine.
00:56
Here we get ch3 cl plus hcl.
01:01
Now further, this ch3 cl again reacts with chlorine.
01:07
And here we get ch2 cl2 plus hcl.
01:16
Now further here we can say ch4 reacts with 2 cl2 and here we get ch2 cl2 plus 2hcl so therefore the balanced equation is when methane one mole of reacts with 2 moles of chlorine.
01:52
Here we get ch2 cl2 plus 2hcl.
02:01
So this is methane.
02:05
Here we have chlorine and the product is dry chloro methane plus hydrochloric acid.
02:25
So this is the final answer.
02:32
Now further for part 2, we need to find mass of chlorine.
02:38
So here we can say, according to equation, here we can say that 2 multiplied by 70 .9 grams of chlorine reacts with 16 .04 grams per mole of methane.
03:15
So this will be equal to 141 .8 grams.
03:24
So here we can say therefore 16 .04 grams of methane reacts with 141 .8 grams of chlorine...