Question

When $ 10ml $ of $ {H_2} $ and $ 12.5ml $ of $ C{l_2} $ are allowed to react, the mixture contains under the same conditions.

Answer 1

Hint :in this particular reaction we have to assume that the reaction is taking place under standard conditions i.e. at standard temperature and pressure. If not, this question has two ways to be solved: first one is considering STP and the second method is just by looking at the values given in $ ml $ .

Complete Step By Step Answer:
At STP conditions, $ 10ml $ of hydrogen will correspond to $ 0.45 $ millimoles and $ 12.5ml $ of chlorine will correspond to $ 0.55 $ millimoles respectively.
The reaction that will take place here is: $ {H_2} + C{l_2} \to 2HCl $
As we can see in the above reaction that 1 mole of hydrogen and 1 mole of Chlorine react with each other to give 2 moles of HCl.
And so $ 0.45 $ millimoles of hydrogen will react with $ 0.45 $ millimoles of Chlorine and will give $ 0.9 $ millimoles of $ HCl $ .
So the amount of Chlorine that remains unreacted in the reaction is $ 0.1 $ millimoles.
Therefore, the reaction mixture will contain $ 0.10 $ millimoles of chlorine and $ 0.9 $ millimoles of HCl.
And hence at STP this will correspond to $ 20.16{\text{ }}ml $ of HCl and $ 2.24{\text{ }}ml $ of Chlorine.

Note :
This problem can be solved alternately also. For solving it in a different way we do not consider any STP, by using the $ ml $ of reactant given we solve this problem.
The reaction that will take place here is: $ {H_2} + C{l_2} \to 2HCl $
And we are given with $ 10ml $ of hydrogen and $ 12.5ml $ of chlorine
 $ 1ml $ Of $ {H_2} $ will react with $ 1ml $ of $ C{l_2} $ to give $ 2ml $ of $ HCl $ .
And so $ 10ml $ of $ {H_2} $ will react with $ 10ml $ of $ C{l_2} $ to give $ 20ml $ of $ HCl $ .
Therefore the amount of chlorine that is left in the mixture:
 $ = 12.5ml - 10ml \\
   = 2.5ml \\ $
And the total amount of $ HCl $ formed is $ 20ml $ .
As you can see , solving with this method also gets almost the same values as that of the first method. There is not much difference so this is also acceptable. To if one needs to find it in a simple way this method can be taken into consideration.